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Solving for a function, not a number
1,000 rupees saved at 5 per cent a year, one stack a year. Each stack is taller than the last, by more each year.
the steps: interest added once a year
close up at year 10
Interest added once a year climbs in steps, to 1,628.89 rupees after 10 years. Added at every moment, it climbs smoothly to 1,648.72.
Put 1,000 rupees in a savings account. The bank adds interest on the money in the account. So the more money there is, the more it earns. In Class 8 you worked out compound interest year by year. Now say the bank adds interest every day, or even every moment. Then the money grows at a rate set by the amount there now.
Two cups of the same tea. The one just poured steams hard. The one poured earlier, nearer the warmth of the room, barely steams.
Tea at 90 °C, room at 30 °C. At first it cools 6 degrees a minute; ten minutes on, at 52.1 °C, only 2.2.
A cup of hot tea on the table works the other way. A very hot cup cools fast. A cup only a little warmer than the room cools slowly. So the tea cools at a rate set by how hot it is now.
Why integrating is not enough
In Chapter 7 you found the water in the roof tank from its flow, a known function of time. Here the rate depends on the amount itself. You cannot just integrate the rate, because it holds the very thing you are looking for.
Here is the question to keep in mind until the last page. When a rate depends on the amount, how do you find the amount?
Left: 2x = 6 has one answer, 3. Right: dy/dx = 2x has a family of answers, the curves y = x² + C.
An equation that holds an unknown function and its derivative is a DIFFERENTIAL EQUATION. Its answer is a function, not a number.
dy/dx = 2x drawn as short slope marks: at every point, a slope mark with slope 2x. Which curves run along the slope marks?
Slope marks show the slope 2x. Three curves y equals x squared plus C follow them. The dot at (1, 2) picks C equals 1.
Start with a simple one, where the rate depends on $x$ alone: $d y/d x = 2 x$. Every curve $y = x^2 + C$ fits it, because the derivative of $x^2 + C$ is $2 x$ for every number $C$. In the figure, the short slope marks show the slope $2 x$ at each point. The curves for $C = 0$, 1 and 2 follow the slope marks. A given value picks one curve. Say $y = 2$ when $x = 1$. Put both into $y = x^2 + C$:
$2 = 1^2 + C$.
Take 1 from both sides:
$C = 1$, so the curve is $y = x^2 + 1$.
The bank account and the cup of tea come back near the end. By then you can solve both.
A SCHOLAR INDIA REMEMBERS
Brahmagupta was an astronomer at Bhillamala, now Bhinmal in Rajasthan, around 628 CE. His tradition worked out the sky from rules, then checked the rules against one careful reading. Ask the same question of the cup of tea: does one reading tell you which curve you are on? Knowing it cools faster when it is hotter does not give you one curve. It gives a family, one for each starting heat. One reading picks your cup’s curve, as $y = 2$ at $x = 1$ picked $C = 1$.
Brahmagupta · the brahminy kite
Check yourself
What makes solving a differential equation different from solving an equation like $x^2 - 4 = 0$?
differential equations always have infinitely many unrelated solutions
the unknown is a whole function, or a family of curves, not a single number
a differential equation can never be checked by substituting back
differential equations only describe physical processes, never pure mathematics
Check your answer
differential equations always have infinitely many unrelated solutions — The solutions in a family are not unrelated, every curve $y=x^2+C$ shares the same shape, shifted up or down by the constant $C$.
✓ the unknown is a whole function, or a family of curves, not a single number — (B) The unknown in a differential equation is a function, not a number, the answer $y=x^2+C$ is a whole family of curves, one per value of $C$, each checked by substituting it back into the original equation.
a differential equation can never be checked by substituting back — Substitution is exactly how the opening example confirms its answer: differentiating $y=x^2+C$ back gives $2 \cdot x$ again, for any $C$.
differential equations only describe physical processes, never pure mathematics — The equations and the methods for solving them are pure mathematics; physical examples like growth and decay are just one use for them.
Substituting $y = x^2 + C$ back into $d y/d x = 2 \cdot x$ confirms the equation holds for \_\_\_\_\_\_.
exactly one value of $C$, fixed by the equation itself
any value of the constant $C$
only positive values of $C$
only the value $C=0$
Check your answer
exactly one value of $C$, fixed by the equation itself — Pinning $C$ to one value is what a PARTICULAR solution does, using an extra condition, this equation alone leaves $C$ free.
✓ any value of the constant $C$ — (B) Differentiating $x^2+C$ gives $2 \cdot x$ regardless of $C$, since a constant’s derivative is always zero, the equation holds for every value of $C$, not a restricted or single one.
only positive values of $C$ — Nothing about $d y/d x = 2 \cdot x$ restricts $C$ to positive numbers; the derivative of $x^2+C$ is $2 \cdot x$ for every real $C$.
only the value $C=0$ — $C$ drops out entirely when you differentiate $x^2+C$, the equation holds no matter which value $C$ takes, not just zero.
Why is the solution to $d y/d x = 2 \cdot x$ called a FAMILY of curves rather than a single curve?
because $2 \cdot x$ can be factored in more than one way
because differentiation always produces more than one answer
because the equation has two different correct answers, $x^2$ and $C$
because infinitely many parabolas satisfy it, one per value of $C$
Check your answer
because $2 \cdot x$ can be factored in more than one way — The right side, $2 \cdot x$, factors however it likes, that has no bearing on why the solution is a family. The family comes from the constant left undetermined by integration.
because differentiation always produces more than one answer — Differentiating $x^2+C$ gives one answer, $2 \cdot x$, every time. It is integrating $2 \cdot x$ back up that leaves an arbitrary constant $C$ open.
because the equation has two different correct answers, $x^2$ and $C$ — $x^2$ and $C$ are not rival answers competing for the same slot, together they form the single general solution $y=x^2+C$, with $C$ free to vary.
✓ because infinitely many parabolas satisfy it, one per value of $C$ — (D) Integrating $2 \cdot x$ back up leaves an arbitrary constant $C$ undetermined, so every value of $C$ gives a different parabola satisfying the same equation, a family, not one curve.
The ORDER of a differential equation is the order of the highest derivative in it. The DEGREE is the highest power of that highest derivative, once the equation is a polynomial in its derivatives. A degree exists only when the equation is a polynomial in every derivative it holds, not only the highest.
(d²y/dx²)³
deg 3
(dy/dx)²
−y
order 2, degree 3
Each term sorted by its derivative. The highest is the second derivative, so the order is 2. It is cubed, so the degree is 3.
Let us read both from one equation:
$((d^2 y)/(d x^2))^3 + ((d y)/(d x))^2 - y = 0$.
The highest derivative is $(d^2 y)/(d x^2)$. So the order is 2. That derivative is raised to the power 3. So the degree is 3. The first derivative is squared, but that does not count. Only the highest derivative sets the degree.
dy/dx
deg 1
sin(dy/dx)
order 1, degree none
Here the derivative sits inside a sine, and a sine is not a power. The order is 1; the degree is not defined.
Some equations have no degree. In $d y/d x + \sin((d y)/(d x)) = 0$ the order is 1. But the derivative sits inside a sine, and a sine is not a polynomial. So the degree is not defined.
d²y/dx²
deg 1
order 2, degree none
(d²y/dx²)²
deg 2
order 2, degree 2
A derivative under a square root. Square both sides first. Then the second derivative is squared: the order is 2, and the degree is 2.
A root needs one more step. If a derivative sits under a root or in a denominator, clear it first: square both sides, or multiply through. Then read the degree.
Read every equation the same way. First find the highest derivative. Its order is the order of the equation. Then clear any root or fraction, and ask: is the equation now a polynomial in all its derivatives? If yes, the degree is the power of the highest derivative. If no, the degree is not defined.
Check each equation in this table against the rule, then cover the last three columns and try it yourself.
Equation
Order
Degree
Why
$d y/d x + y = x$
1
1
The highest derivative is the first, to the power 1.
$(d^2 y)/(d x^2) + y = 0$
2
1
The second derivative, to the power 1. The next figure’s curves fit it.
$(d^2 y)/(d x^2) + ((d y)/(d x))^3 + y = 0$
2
1
The cube is on the first derivative, the lower one. It does not count.
$(d^2 y)/(d x^2) = \sqrt{1 + (d y)/(d x)}$
2
2
Square both sides first: $((d^2 y)/(d x^2))^2 = 1 + d y/d x$.
$d y/d x + 1/((d y)/(d x)) = x$
1
2
Multiply through by $d y/d x$ first: $((d y)/(d x))^2 + 1 = x d y/d x$.
$(d^2 y)/(d x^2) + \sin((d y)/(d x)) = 0$
2
not defined
A derivative sits inside a sine, and a sine is not a polynomial.
$(d^2 y)/(d x^2) + e^{(d y)/(d x)} = 0$
2
not defined
A derivative sits in a power of $e$, which is not a polynomial.
Four members of y equals A cos x plus B sin x. All four fit one equation, of order 2 and degree 1.
The figure draws four curves from the family $y = A \cos x + B \sin x$. All four fit one equation, $(d^2 y)/(d x^2) + y = 0$. Its order is 2 and its degree is 1.
Your turn: find the order and degree of $d y/d x = \sqrt{1 + ((d^2 y)/(d x^2))^3}$.
Check yourself
What does the ORDER of a differential equation measure?
the highest power any variable is raised to, ignoring derivatives
the order of the highest derivative appearing in it
the number of arbitrary constants in its general solution
the number of terms written in the equation
Check your answer
the highest power any variable is raised to, ignoring derivatives — Order is specifically about the highest DERIVATIVE’s order, not about ordinary powers of $x$ or $y$ with no derivative attached.
✓ the order of the highest derivative appearing in it — (B) Order is defined directly as the order of the highest derivative in the equation , term count, constant count, and ordinary powers are each a different, unrelated fact.
the number of arbitrary constants in its general solution — The number of constants in the general solution matches the order, but that is a downstream fact ABOUT order, order itself is defined by the highest derivative.
the number of terms written in the equation — An equation could be rewritten with more or fewer terms without changing its order, order depends only on the highest derivative present.
For $((d^2 y)/(d x^2))^3 + ((d y)/(d x))^2 - y = 0$, what are its order and degree?
order 3, degree 2
order 2, degree 2
order 2, degree 3
order 2, degree undefined
Check your answer
order 3, degree 2 — Order and degree are swapped here: the highest derivative, $(d^2 y)/(d x^2)$, sets order at 2, and ITS power, 3, is the degree, not the other way round.
order 2, degree 2 — Degree comes from the power on the HIGHEST-order derivative only. The power 2 belongs to the lower-order term $((d y)/(d x))^2$, which does not decide degree.
✓ order 2, degree 3 — (C) The highest derivative is $(d^2 y)/(d x^2)$, giving order 2; its own power in the equation is 3, giving degree 3, the lower-order term’s power plays no part in degree.
order 2, degree undefined — Every term here is a plain polynomial power of a derivative, no derivative sits inside a function like $\sin$, so degree is defined, and equals 3.
For $(d^3 y)/(d x^3) + ((d y)/(d x))^4 = x$, what are its order and degree?
order 3, degree 4
order 4, degree 3
order 3, degree 1
order 3, degree 3
Check your answer
order 3, degree 4 — $4$ is the power on $d y/d x$, a LOWER-order term. Degree only ever reads the power on the highest-order derivative, $(d^3 y)/(d x^3)$, which appears to the first power.
order 4, degree 3 — The highest derivative present is $(d^3 y)/(d x^3)$, third order, nothing in the equation reaches a fourth derivative.
✓ order 3, degree 1 — (C) The highest derivative is $(d^3 y)/(d x^3)$, order 3, and it appears to the first power, degree 1, regardless of the fourth power sitting on the lower-order term $d y/d x$.
order 3, degree 3 — The highest-order derivative, $(d^3 y)/(d x^3)$, appears with no exponent at all, which means power 1, not 3.
For $d y/d x + \sin(x) = 0$ (note $\sin$ of $x$ itself, not of the derivative), what is its degree?
undefined, since a trig function appears in the equation
2
1
0
Check your answer
undefined, since a trig function appears in the equation — The genuine trap equation is $d y/d x + \sin((d y)/(d x))=0$, where $\sin$ wraps the DERIVATIVE. Here $\sin(x)$ wraps only $x$, leaving $d y/d x$ a plain polynomial term, degree stays defined.
2 — No term here is raised to the second power, $d y/d x$ appears to the first power only, giving degree 1, not 2.
✓ 1 — (C) $\sin(x)$ wraps $x$, not the derivative, $d y/d x$ still appears as a plain first-power term, so the equation is polynomial in its derivative and degree is defined, equal to 1.
0 — $d y/d x$ appears plainly in the equation, to the first power, the derivative is very much present, so degree cannot be $0$.
✕MISCONCEPTION
Step 1
dy/dx + sin(dy/dx) = 0
order 1
Step 2
degree 1
?
The working for dy/dx + sin(dy/dx) = 0, in two steps. Which step is wrong? Keep your answer.
Here is a working for $d y/d x + \sin((d y)/(d x)) = 0$. Which step is wrong?
The highest derivative is $d y/d x$, so the order is 1.
$d y/d x$ appears to the power 1, so the degree is 1.
Keep your answer, then look again at the equation. The first step is right. But the derivative also sits inside a sine, and a sine of $d y/d x$ is not a power of it. So the equation is not a polynomial in its derivatives. It has no degree, and the second step is wrong.
Step 1
dy/dx + sin(dy/dx) = 0
order 1
Step 2
no degree
Step 1 stands: the order is 1. Step 2 is wrong. The derivative sits inside a sine, so there is no degree.
d²y/dx²
deg 1
sin(dy/dx)
order 2, degree none
The second derivative carries power 1, but the first sits inside a sine. The order is 2, and there is no degree.
A lower derivative inside a sine does the same. In $(d^2 y)/(d x^2) + \sin((d y)/(d x)) = 0$, the second derivative has the power 1. Still, the equation has no degree. It must be a polynomial in every derivative it holds.
(dy/dx)²
deg 2
y
order 1, degree 2
d²y/dx²
deg 1
y
order 2, degree 1
Order and degree come apart: (dy/dx)² + y = 0 is order 1, degree 2; d²y/dx² + y = 0 is order 2, degree 1.
Every differential equation has an order. Not every one has a degree.
Check yourself
A student claims order and degree measure “the same thing”: just how many derivatives an equation has. What is wrong with this?
degree counts the highest derivative, and order is its power instead
order is the highest derivative present; degree is its own power
nothing is wrong; order and degree really do always give the same number
order and degree only differ for equations with more than one dependent variable
Check your answer
degree counts the highest derivative, and order is its power instead — This just swaps which definition belongs to which word. Order is the highest derivative’s own ORDER; degree is that same derivative’s POWER.
✓ order is the highest derivative present; degree is its own power — (B) Order and degree measure two different things, which derivative is highest, and that derivative’s own power, and the second one additionally needs the equation to be polynomial in its derivatives.
nothing is wrong; order and degree really do always give the same number — Order and degree commonly differ, for $((d^2 y)/(d x^2))^3 + ((d y)/(d x))^2 - y=0$, order is 2 while degree is 3.
order and degree only differ for equations with more than one dependent variable — Every equation in this chapter has one dependent variable, $y$, yet order and degree still differ whenever the highest-order derivative’s power is not 1.
What is the order and degree of $d y/d x + \sin((d y)/(d x)) = 0$?
order 0, degree equal to 1
order 1, degree equal to 0
order 1, no degree defined
order 1, degree equal to 1
Check your answer
order 0, degree equal to 1 — $d y/d x$ is a first derivative, plainly present, order is 1, not 0.
order 1, degree equal to 0 — “Undefined” and “$0$” are not the same thing. A degree of $0$ would mean the derivative does not appear at all, which is false here, it appears inside $\sin$, where no polynomial power can be read.
✓ order 1, no degree defined — (C) The highest derivative present is $d y/d x$, a first derivative, so order is 1, but it sits inside $\sin(\ldots)$, a non-polynomial expression, so degree has no value to report at all.
order 1, degree equal to 1 — $\sin((d y)/(d x))$ is not a polynomial term in $d y/d x$, so no power can be read off it, degree is undefined here, even though order is perfectly fine.
Why is the degree undefined for $d y/d x + \sin((d y)/(d x)) = 0$?
because the equation has no solution at all
because $\sin$ is not defined for derivatives of any kind
$\sin((d y)/(d x))$ is not a polynomial term
because the order is also undefined in this equation
Check your answer
because the equation has no solution at all — Degree being undefined is a labeling issue about polynomial form, not a statement about whether the equation can be solved.
because $\sin$ is not defined for derivatives of any kind — $\sin$ takes $d y/d x$ as its input just as happily as it takes any other quantity, the issue is that $\sin((d y)/(d x))$ is not a polynomial in $d y/d x$, not that $\sin$ refuses the input.
✓ $\sin((d y)/(d x))$ is not a polynomial term — (C) $\sin((d y)/(d x))$ is not a polynomial term in the derivative, which is exactly the condition degree needs, order, by contrast, only needs the highest derivative named, which stays perfectly clear at 1.
because the order is also undefined in this equation — Order only needs the highest derivative identified, which is $d y/d x$ here, order is 1, and stays perfectly well defined.
Chapter 7’s roof tank had a family of answers, one for each starting level. The level at one moment picked one. A differential equation works the same way.
A starting value, left, y = 0.5 at x = 0, picks C = 0.5. A point, right, (2, 3.5), picks C = 1.5.
Its GENERAL SOLUTION is the whole family, with its ARBITRARY CONSTANTS still free. A PARTICULAR SOLUTION is one member of the family. You get it by fixing each constant from a given fact. The fact may be a starting value, or a point the curve passes through.
y = 1 at x = 0
and y = 3 at x = 2
The second derivative is 0, order 2: y(0) = 1 leaves a fan of lines; y(2) = 3 picks one, y = 1 + x.
An equation of order $n$ has $n$ arbitrary constants in its general solution. So it needs $n$ given facts to pick one member. An equation of order 1 needs one fact. An equation of order 2 needs two.
Four members of y equals x squared over 2 plus C, 1 apart. Knowing y is 3.5 at x equals 2 picks C equals 1.5.
Let us find one member. The equation $d y/d x = x$ has the general solution $y = x^2/2 + C$. The figure draws four members, for $C = -1.5$, $-0.5$, $0.5$ and $1.5$. They are 1 apart at every $x$. Now say $y = 3.5$ when $x = 2$.
Worked example
Find the member through $(2, 3.5)$
$y = x^2/2 + C$ Start from the general solution.
$3.5 = 2^2/2 + C$ Put $x = 2$ and $y = 3.5$ into it.
$3.5 = 2 + C$ Work out $2^2/2 = 2$.
$C = 1.5$ Take 2 from both sides.
*$y = x^2/2 + 1.5$* Put $C = 1.5$ back: this is the particular solution, the top curve.
Check: $x = 2$ gives $y = 2 + 1.5 = 3.5$ The curve passes through the given point.
The given fact is not part of the equation. So a question that asks for a particular solution must give you one fact for each constant.
Your turn: the general solution of an equation is $y = A \cos x + B \sin x$. How many given facts does it take to pick one member?
Check yourself
What distinguishes a general solution of an order-$n$ differential equation from a particular solution?
the general solution has $n$ arbitrary constants; the particular solution has them fixed
the general solution applies to first-order equations only; the particular solution applies to every order
the general solution solves the equation exactly; the particular solution only solves it approximately
the general solution is always a straight line; the particular solution is always a curve
Check your answer
✓ the general solution has $n$ arbitrary constants; the particular solution has them fixed — (A) An order-$n$ equation’s general solution carries $n$ independent arbitrary constants, describing a whole family of curves; a particular solution fixes each constant using given conditions, leaving one specific curve.
the general solution applies to first-order equations only; the particular solution applies to every order — General and particular solutions exist for equations of every order, an order-$n$ equation’s general solution simply carries $n$ constants.
the general solution solves the equation exactly; the particular solution only solves it approximately — Both solutions satisfy the equation exactly when substituted back, the difference is purely in how many constants remain free, not in precision.
the general solution is always a straight line; the particular solution is always a curve — Nothing forces a general solution to be a straight line, it is a family of curves shaped however the equation dictates, such as the parabolas $y=x^2+C$.
An order-3 differential equation’s general solution contains how many independent arbitrary constants?
2
3
1
a number that depends on which particular solution is chosen
Check your answer
2 — Two arbitrary constants matches an order-2 equation. Order 3 needs one more constant still, three in total.
✓ 3 — (B) An order-3 equation’s general solution carries exactly three independent arbitrary constants, the count is fixed by the order itself, whatever particular solution is later chosen.
1 — One arbitrary constant matches an order-1 equation. An order-3 equation’s general solution needs three.
a number that depends on which particular solution is chosen — The NUMBER of constants is fixed by the order, always three here, choosing a particular solution only fixes their VALUES, not how many there are.
What information turns a general solution into a particular one?
computing one further derivative of the general solution
rewriting the equation in variables-separable form
given conditions, such as an initial value
raising the order of the original differential equation
Check your answer
computing one further derivative of the general solution — Differentiating the general solution again gives information about a higher derivative, not a fixed value for the existing constants.
rewriting the equation in variables-separable form — Separable form is about how to SOLVE an equation, not about how to pin down a family’s constants once a solution is already found.
✓ given conditions, such as an initial value — (C) Given conditions, an initial value, or a specific point the curve must pass through , supply exactly enough information to fix every constant, turning the general family into one particular curve.
raising the order of the original differential equation — Raising the order would introduce even more arbitrary constants to determine, the opposite of what fixing a particular solution does.
Forming the equation from a family (beyond the board syllabus)
▲CONCEPT
So far you started from an equation and found a family. Here the work runs the other way. You start from a family of curves and find the equation they all fit.
Eliminating the constants
A family with $n$ arbitrary constants fits an equation of order $n$. To find it, you ELIMINATE the constants. Differentiate the family’s equation $n$ times. That gives $n$ new equations. Combine them with the first one, so that every constant drops out. What is left holds $x$, $y$ and the derivatives, and no constant.
The lines y = mx for five values of m. Each runs along the slope marks of slope y/x, so each fits x dy/dx = y.
Let us try a family with one constant: the lines $y = m x$ through the origin.
Worked example
Form the equation of the lines $y = m x$
$y = m x$ There is one constant, $m$, so differentiate once to remove it.
$d y/d x = m$ Differentiate both sides with respect to $x$.
$m = y/x$ Solve the first line for $m$, which needs $x \neq 0$.
$d y/d x = y/x$ Put this $m$ into the second line, and $m$ is gone.
*$x d y/d x = y$* Multiply both sides by $x$: one constant gave an equation of order 1.
Check: $x \cdot m = m x = y$ Put $d y/d x = m$ back: every line $y = m x$ fits.
1 constant: order 1
2 constants: order 2
Count the constants. One, in y = mx: order 1. Two, in y = A cos x + B sin x: order 2.
Count the constants first. That count is the order you will reach, and it tells you how many times to differentiate. The worked example below has two constants.
Your turn: form the equation of the family $y = C e^x$. What order do you expect, before you start?
Check yourself
To form the differential equation of a family of curves with 2 arbitrary constants, how many times must you differentiate?
3
2
it depends on the specific family, never on the constant count
1
Check your answer
3 — Two differentiations already supply enough new equations, combined with the original, to eliminate both constants, a third is not needed.
✓ 2 — (B) Two arbitrary constants need two new equations to eliminate them alongside the original, so exactly two differentiations are required, the count tracks the constant count directly.
it depends on the specific family, never on the constant count — The number of differentiations needed is fixed exactly by the constant count, here two, it is not left to vary case by case.
1 — One differentiation gives only one new equation, not enough, together with the original, to eliminate two separate constants.
A family of curves has 3 arbitrary constants. What order will its differential equation be?
4
the degree, not the order, matches the constant count
3
2
Check your answer
4 — Order 4 would require a fourth constant to eliminate, which this family does not have, three constants give order 3.
the degree, not the order, matches the constant count — It is the ORDER that matches the constant count directly, since eliminating $n$ constants takes $n$ differentiations, reaching an order-$n$ equation. Degree is a separate fact about the resulting equation.
✓ 3 — (C) Eliminating three arbitrary constants takes three differentiations, and the resulting differential equation has order 3, the order, not the degree, tracks the constant count directly.
2 — Order 2 would come from a family carrying only two constants. Three constants push the order up to three.
After differentiating a 2-constant family twice, why must the results be combined algebraically with the original equation?
to increase the order of the equation further
to check the answer using specific numbers
to eliminate both constants from the system
to convert the equation into variables-separable form
Check your answer
to increase the order of the equation further — The order is already fixed by how many times you differentiated, combining the equations afterward eliminates constants, it does not add any further derivatives.
to check the answer using specific numbers — Plugging in numbers checks a FINISHED equation. Combining the three equations here is how the constants get eliminated in the first place, before any checking happens.
✓ to eliminate both constants from the system — (C) The original equation and its two derivatives together give three equations with the same two constants, enough algebra to eliminate both, leaving one equation in $x$, $y$, and derivatives only.
to convert the equation into variables-separable form — Forming the equation (eliminating constants) and solving it (separating variables, say) are separate tasks, this combining step belongs entirely to forming.
Form the differential equation of the family $y = C x^2$, where $C$ is an arbitrary constant.
$x ((d y)/(d x)) = 2 y$
$x ((d y)/(d x)) + 2 y = 0$
$x ((d y)/(d x)) = y$
$((d y)/(d x)) = 2 y$
Check your answer
✓ $x ((d y)/(d x)) = 2 y$ — (A) Differentiating $y=C x^2$ gives $d y/d x = 2 C x$; substituting $C=y/x^2$ from the original equation gives $d y/d x = 2 y/x$, so $x ((d y)/(d x)) = 2 y$.
$x ((d y)/(d x)) + 2 y = 0$ — There is no sign to flip here: substituting $C=y/x^2$ into $d y/d x=2 C x$ gives $d y/d x=2 y/x$, a plus, not a relation that opposes $y$ and its derivative.
$x ((d y)/(d x)) = y$ — Differentiating $C x^2$ gives $2 C x$, carrying a factor of $2$ that must stay in the eliminated equation, dropping it gives the wrong family.
$((d y)/(d x)) = 2 y$ — Without the factor of $x$, the equation no longer reduces correctly back to $C=y/x^2$, the $x$ is required for the elimination to work.
Left: one member, with a short slope line. Right: the same member and its second derivative, minus y, mirrored in the x-axis.
Worked example
Form the differential equation of $y = A \cos(x) + B \sin(x)$
$y = A \cos(x) + B \sin(x)$, with $A, B$ arbitrary constants There are two constants, $A$ and $B$. So differentiate twice to eliminate them.
$d y/d x = -A \sin(x) + B \cos(x)$ Differentiate once. Both constants are still there.
$(d^2 y)/(d x^2) = -A \cos(x) - B \sin(x)$ Differentiate again.
$(d^2 y)/(d x^2) = -y$ The right side is $-(A \cos(x) + B \sin(x))$, which is $-y$.
*$(d^2 y)/(d x^2) + y = 0$* No $A$ or $B$ is left. The order is 2, one for each constant.
Check: $(-A \cos(x) - B \sin(x)) + (A \cos(x) + B \sin(x)) = 0$ Put $y$ and its second derivative back into the equation. The terms cancel for every $A$ and $B$.
y'' = −y
y = 1.5 cos x + sin x, differentiated twice. The slope still holds both constants; the second derivative is the curve turned over, −y.
Check yourself
In forming the differential equation for $y = A \cos(x) + B \sin(x)$, why are exactly two differentiations needed?
because there are two arbitrary constants, $A$ and $B$, to eliminate
because the final equation must reach order 2, by convention
because $\cos$ and $\sin$ each independently require one derivative to simplify
because differentiating twice is required for any trigonometric function, by convention
Check your answer
✓ because there are two arbitrary constants, $A$ and $B$, to eliminate — (A) Two constants, $A$ and $B$, need eliminating, and eliminating $n$ constants always takes $n$ differentiations, the trigonometric functions themselves play no separate role in that count.
because the final equation must reach order 2, by convention — The order 2 is a RESULT of eliminating two constants, not a target chosen in advance, the reasoning runs from constants to order, not the other way round.
because $\cos$ and $\sin$ each independently require one derivative to simplify — The functions $\cos$ and $\sin$ do not individually demand a derivative each, the differentiation count is set purely by how many constants, $A$ and $B$, need eliminating.
because differentiating twice is required for any trigonometric function, by convention — There is no rule requiring two derivatives for trigonometric functions specifically, the count always tracks the constant count, whatever functions are involved.
For $y = A \cos(x) + B \sin(x)$, what is $(d^2 y)/(d x^2)$ in terms of $y$?
$-2 y$
$-y$
$0$
$y$
Check your answer
$-2 y$ — Differentiating twice returns exactly $-y$, with coefficient $1$ (negated), no doubling occurs anywhere in the two steps.
✓ $-y$ — (B) Differentiating once gives $-A \sin(x) + B \cos(x)$; differentiating again gives $-A \cos(x) - B \sin(x)$, which is exactly $-y$.
$0$ — The second derivative does not vanish, it returns $-A \cos(x) - B \sin(x)$, which is exactly $-y$, not zero.
$y$ — $d y/d x = -A \sin(x) + B \cos(x)$, and differentiating again gives $-A \cos(x) - B \sin(x)$, the negative sign is part of the result, not something to drop.
The resulting equation $(d^2 y)/(d x^2) + y = 0$ has order 2, matching the two eliminated constants. What does this illustrate in general?
order 2 confirms the family only ever had one constant
eliminating $n$ constants always produces an order-$n$ equation
order and degree happen to agree in this particular example
the order could have come out as any number here, and 2 is a coincidence
Check your answer
order 2 confirms the family only ever had one constant — Order 2 reflects the TWO constants, $A$ and $B$, eliminated here, one constant would have produced an order-1 equation instead.
✓ eliminating $n$ constants always produces an order-$n$ equation — (B) This worked example is the general rule in miniature: eliminating $n$ arbitrary constants from a family always produces a differential equation of order exactly $n$.
order and degree happen to agree in this particular example — $(d^2 y)/(d x^2)+y=0$ has degree 1, since the highest derivative appears to the first power, order and degree do not match here at all.
the order could have come out as any number here, and 2 is a coincidence — The order is not left to chance, eliminating $n$ constants always yields an order-$n$ equation, and this worked example is exactly that rule in action.
An equation $d y/d x = g(x) h(y)$ is VARIABLES SEPARABLE. Its right side is a function of $x$ times a function of $y$.
the y side
dy1/h(y)
the x side
g(x)dx
Separating the variables: every piece that holds y goes to one side, and every piece that holds x to the other.
Moving dy and dx
One symbol, not a fraction. $d y/d x$ stands for the derivative of $y$ with respect to $x$. It is not a number $\mathrm{d} y$ divided by a number $\mathrm{d} x$.
This chapter still moves $\mathrm{d} y$ and $\mathrm{d} x$ as if they were pieces. It writes lines like $(\mathrm{d} y)/h(y) = g(x) \mathrm{d} x$. Such a line is short for this: integrate $1/h(y)$ with respect to $y$, integrate $g(x)$ with respect to $x$, and set the two integrals equal.
Why it works
Why separating is allowed
$d y/d x = g(x) h(y)$ Start from the separable equation, where $h(y) \neq 0$.
$1/h(y) \cdot d y/d x = g(x)$ Divide both sides by $h(y)$.
$H'(y) d y/d x = g(x)$ Let $H$ be an integral of $1/h$. Since $H'(y) = 1/h(y)$, that is the left side.
$d/d x (H(y)) = g(x)$ By the chain rule, the derivative of $H(y)$ with respect to $x$ is $H'(y) d y/d x$: the line above.
$H(y) = G(x) + C$ Integrate both sides with respect to $x$. Let $G$ be an integral of $g$.
The shorthand gives the same line, so the steps below use it. First, multiply both sides of $d y/d x = g(x) h(y)$ by $\mathrm{d} x$:
$\mathrm{d} y = g(x) h(y) \mathrm{d} x$.
Then divide both sides by $h(y)$, where $h(y) \neq 0$. Now every $y$ is on the left and every $x$ on the right:
$(\mathrm{d} y)/h(y) = g(x) \mathrm{d} x$.
Integrate both sides, the left with respect to $y$ and the right with respect to $x$. The two constants combine into one $+ C$ on one side:
$\int (\mathrm{d} y)/h(y) = \int g(x) \mathrm{d} x + C$.
Each side is an integral from Chapter 7.
the y side
dy1/y²
the x side
cos xdx
For dy/dx = y² cos x, y ≠ 0: dy and 1/y² go to the y side, cos x and dx to the x side.
The slope marks of dy/dx = y² cos x, and y = −1/(sin x + C) for C = −3, −2, 2, 3, running along them.
Let us solve $d y/d x = y^2 \cos x$. Here $g(x) = \cos x$ and $h(y) = y^2$.
Worked example
Separate, integrate, check
$\mathrm{d} y = y^2 \cos x \mathrm{d} x$ Multiply both sides by $\mathrm{d} x$.
$(\mathrm{d} y)/y^2 = \cos x \mathrm{d} x$ Divide both sides by $y^2$, which needs $y \neq 0$.
$\int (\mathrm{d} y)/y^2 = \int \cos x \mathrm{d} x$ Integrate both sides, each with respect to its own variable.
$-1/y = \sin x + C$ The integral of $y^{-2}$ is $-1/y$, and one $+ C$ covers both sides.
*$y = -1/(\sin x + C)$* Solve for $y$: turn both sides upside down and change the sign.
Check: $d y/d x = \cos x/(\sin x + C)^2 = y^2 \cos x$ Differentiate the answer and use $y^2 = 1/(\sin x + C)^2$.
The line y = 0 fits too: every slope mark on the x-axis is flat. Dividing by y² lost it.
The division needs care. The line $y = 0$ also fits the equation, since both sides are then 0. Dividing by $y^2$ lost it. So say where you divide.
Your turn: $d y/d x = e^{x + y}$ and $d y/d x = e^x + y$ look alike. Which one separates?
A SCHOLAR INDIA REMEMBERS
Was anything lost when you divided by $y^2$? Check $y = 0$: both sides of $d y/d x = y^2 \cos x$ are then $0$, so it is a solution. In the family $y = -1/(\sin x + C)$, a fraction with $-1$ on top is never $0$, for any $C$, so the family never reaches it. Record $y = 0$ separately, or it is lost.
Brahmagupta · the brahminy kite
Check yourself
A first-order ODE is called VARIABLES SEPARABLE when \_\_\_\_\_\_.
it has exactly one arbitrary constant in its solution
the right side is a sum of a function of $x$ and a function of $y$
$y$-terms move to one side, $x$-terms to the other
it contains no product of $x$ and $y$
Check your answer
it has exactly one arbitrary constant in its solution — Having one arbitrary constant is true of every first-order equation’s general solution, separable or not, it does not single out the separable class.
the right side is a sum of a function of $x$ and a function of $y$ — A sum is not enough. $d y/d x = x + y$ is a sum of a function of $x$ and a function of $y$, and its variables cannot be separated: separable needs the product form $g(x) h(y)$.
✓ $y$-terms move to one side, $x$-terms to the other — (C) Variables separable means the equation can be rearranged so every $y$-term sits on one side and every $x$-term on the other, a description of the equation’s SHAPE, not of its constant count or any one example.
it contains no product of $x$ and $y$ — Having no product of $x$ and $y$ is not enough. $d y/d x = x + y$ has none, and it is not separable. The test is whether the right side can be written as $g(x) h(y)$.
After separating $d y/h(y) = g(x) d x$, what is the next step to find the general solution?
multiply through by the integrating factor
integrate both sides
substitute $y = v x$
differentiate both sides once more
Check your answer
multiply through by the integrating factor — An integrating factor is the tool for LINEAR equations that resist separation entirely, a separated equation needs no multiplier at all, only integration.
✓ integrate both sides — (B) Once the equation is separated into $d y/h(y) = g(x) d x$, integrating both sides directly gives the general solution, no substitution or multiplier is needed at this stage.
substitute $y = v x$ — The substitution $y=v x$ is how a HOMOGENEOUS equation gets turned into separated form in the first place, it plays no role once separation is already done.
differentiate both sides once more — Differentiating again raises the order and moves away from a solution, separating is meant to be followed by integrating, the reverse operation.
Which of these first-order equations is variables separable?
$d y/d x = (x+y)/x$
$d y/d x = x + y$
$d y/d x + y = x$
$d y/d x = x^2 y$
Check your answer
$d y/d x = (x+y)/x$ — Dividing $(x+y)$ by $x$ gives $1+y/x$, a function of $y/x$ alone, that is the signature of a HOMOGENEOUS equation, solved by substituting $y=v x$ first, not by separating directly.
$d y/d x = x + y$ — $x+y$ on the right adds $x$ and $y$ together with no multiplication or division separating them, there is no algebraic move that isolates $y$-terms from $x$-terms here.
$d y/d x + y = x$ — $d y/d x+y=x$ fits the linear form $d y/d x+P(x)y=Q(x)$ with $P(x)=1$. Its $x$ and $y$ terms cannot simply be moved to opposite sides.
✓ $d y/d x = x^2 y$ — (D) $d y/d x = x^2 y$ rearranges directly into $d y/y = x^2 d x$ by ordinary algebra, the other three each need a substitution or an integrating factor before anything can be separated.
Slope marks of dy/dx = x/y, with y² − x² = K for four K. The point (1.5, 2.5) picks K = 4.
y squared minus x squared equals K, for K equals 1, 2, 4 and 6. At (1.5, 2.5) the slope is x over y: 0.6.
Worked example
Solve $d y/d x = x/y$ by separating variables
$d y/d x = x/y$ The right side is $x$ times $1/y$, so the variables can be separated. The equation needs $y \neq 0$.
$y \mathrm{d} y = x \mathrm{d} x$ Multiply both sides by $y \mathrm{d} x$. Every $y$ is now on the left, and every $x$ on the right.
$y^2/2 = x^2/2 + C$ Integrate both sides. One $+ C$ goes on one side.
$y^2 - x^2 = K$ Multiply by 2, and write $K = 2 C$. This is the general solution.
$2 \cdot y \cdot d y/d x - 2 \cdot x = 0$ Check the answer. Differentiate $y^2 - x^2 = K$ with respect to $x$.
*$d y/d x = x/y$* Solve for $d y/d x$. You get the original equation back.
Check yourself
Solving $d y/d x = x/y$ by separating variables, which is the rearrangement with the variables separated: every $y$ with $d y$ on one side, every $x$ with $d x$ on the other?
$y d y = x d x$
$d y/y = x d x$
$y d x = x d y$
$d y/y = d x/x$
Check your answer
✓ $y d y = x d x$ — (A) Cross-multiplying $d y/d x=x/y$ gives $y d y = x d x$, with $y$ paired to $d y$ and $x$ paired to $d x$, ready to integrate directly.
$d y/y = x d x$ — $d y/y = x d x$ separates $d y/d x = x y$, a product. Here $x/y$ is a quotient: multiply both sides by $y$ and by $d x$ to get $y d y = x d x$.
$y d x = x d y$ — Cross-multiplying $d y/d x=x/y$ gives $y d y = x d x$, $y$ pairs with $d y$ and $x$ pairs with $d x$, not swapped across.
$d y/y = d x/x$ — $d y/y = d x/x$ is the separation of $d y/d x = y/x$, not of this equation. Here, cross-multiplying $d y/d x=x/y$ gives $y$ paired with $d y$, not $y$ dividing $d y$.
Solve $d y/d x = -x/y$ by separating variables. What is the general solution?
$x^2 - y^2 = K$
$x^2 + y^2 = K$
$x + y = K$
$y^2 - x^2 = K$
Check your answer
$x^2 - y^2 = K$ — Separating $d y/d x=-x/y$ gives $y d y=-x d x$; integrating gives $y^2/2=-x^2/2+C$, which combines to $x^2+y^2=K$, both terms add, they do not subtract.
✓ $x^2 + y^2 = K$ — (B) Separating gives $y d y=-x d x$; integrating gives $y^2/2=-x^2/2+C$, which rearranges to $x^2+y^2=K$, a family of circles, unlike the hyperbolas from the original, unflipped equation.
$x + y = K$ — Integrating $y d y=-x d x$ produces squared terms, $y^2/2$ and $x^2/2$, there is no step that removes the squares to leave a linear family.
$y^2 - x^2 = K$ — $y^2-x^2=K$ solves the ORIGINAL equation $d y/d x=x/y$. Flipping the sign to $d y/d x=-x/y$ changes the family entirely, to $x^2+y^2=K$, a circle.
How does verifying $y^2 - x^2 = K$ solves $d y/d x = x/y$ actually work?
confirm the family has exactly one arbitrary constant
differentiate the solution implicitly and check it reduces back to the original equation
confirm that $K$ comes out to a whole number
substitute specific numbers for $x$ and $y$ and check the equation holds for that one pair
Check your answer
confirm the family has exactly one arbitrary constant — Every first-order equation’s general solution carries one constant, whether or not it actually solves the equation, constant count alone verifies nothing.
✓ differentiate the solution implicitly and check it reduces back to the original equation — (B) Differentiating $y^2-x^2=K$ implicitly gives $2 y ((d y)/(d x)) - 2 x = 0$, so $d y/d x = x/y$, exactly the original equation, confirmed for the whole family at once, not just one point.
confirm that $K$ comes out to a whole number — $K$ can be any real number, not just a whole number, its value depends on which member of the family is chosen, and has no bearing on correctness.
substitute specific numbers for $x$ and $y$ and check the equation holds for that one pair — One matching pair of numbers only shows that point lies on the curve, verifying the SOLUTION means checking the relationship holds along the whole family, which implicit differentiation does directly.
The slope marks of dy/dx = 1 + y/x. On each ray from the origin y/x is fixed, so every slope mark along one ray is parallel.
An equation $d y/d x = F(x, y)$ is HOMOGENEOUS when $F(\lambda x, \lambda y) = F(x, y)$ for every number $\lambda \neq 0$. Then $F$ depends only on the ratio $y/x$.
Always test this first. For $F(x, y) = (x + y)/x$, $F(\lambda x, \lambda y) = (\lambda x + \lambda y)/(\lambda x) = (x + y)/x$. So this equation is homogeneous.
Substituting for the ratio
Then put $y = v x$, where $v$ is a new function of $x$:
$y = v x$.
Differentiate both sides with respect to $x$. The right side is $v$ times $x$, so use the product rule. The derivative of $x$ is 1:
$d y/d x = v \cdot 1 + x d v/d x = v + x d v/d x$.
Put both into the equation. It becomes one in $v$ and $x$, and it separates. Solve it, then put $v = y/x$ back. An answer still in $v$ is not finished.
Six rays from the origin. Along the ray y equals 2x, y over x is 2, so all three points get slope 3.
The figure shows why the ratio is enough. Each ray from the origin has one value of $y/x$. On the ray $y = 2 x$ the ratio is 2 at every point. So $d y/d x = 1 + y/x$ gives the same slope, 3, at all three marked points.
Some equations are written the other way, as $d x/d y = h(x/y)$. For those, put $x = v y$ instead.
Your turn: is $d y/d x = (x + y + 1)/x$ homogeneous? Test it with $\lambda$.
Check yourself
A first-order ODE $d y/d x = F(x,y)$ is HOMOGENEOUS when \_\_\_\_\_\_.
the equation is already separable in $x$ and $y$
$F(x,y)$ can be written as a function of $y/x$ alone
$F(x,y)$ contains only first powers of $x$ and $y$
the family of solution curves must pass through the origin
Check your answer
the equation is already separable in $x$ and $y$ — A homogeneous equation is usually not separable as written, the whole point of the $y=v x$ substitution is to MAKE it separable, which means it was not separable to begin with.
✓ $F(x,y)$ can be written as a function of $y/x$ alone — (B) $F(x,y)$ reducing to a function of $y/x$ alone is the whole test for homogeneity , not separability, not restricted powers, and not any geometric fact about the solution curves.
$F(x,y)$ contains only first powers of $x$ and $y$ — Requiring only first powers is far too narrow, $(x^2+y^2)/(x y)$ also reduces to a function of $y/x$ alone and counts as homogeneous, despite its squared terms.
the family of solution curves must pass through the origin — Whether solution curves pass through the origin is not part of the definition at all, homogeneity is a statement about $F(x,y)$’s form, not about where any particular curve sits.
For $d y/d x = g(y/x)$, which substitution gives an equation in $v$ and $x$ with its variables separable?
$y = v x$
$y = v x^2$
$v = x y$
$y = v + x$
Check your answer
✓ $y = v x$ — (A) Putting $y = v x$, so $v = y/x$, turns $g(y/x)$ into $g(v)$, and the equation becomes $x d v/d x = g(v) - v$, an equation in $v$ and $x$ whose variables separate.
$y = v x^2$ — $y = v x^2$ makes $y/x = v x$, so $g(y/x)$ becomes $g(v x)$, still a function of both $v$ and $x$. Put $y = v x$, so that $v = y/x$.
$v = x y$ — $v=x y$ does not capture the $y/x$ structure that makes the equation homogeneous, the substitution needs $y$ expressed as $v$ TIMES $x$, so that $v=y/x$.
$y = v + x$ — The substitution is multiplicative, $y=v x$, not additive, adding $v$ and $x$ does not turn $F(x,y)$ into a function of $v$ alone.
After substituting $y = v x$, what is $d y/d x$ in terms of $v$ and $x$?
$v ((d v)/(d x))$
$(v) + ((d v)/(d x))$
$x ((d v)/(d x))$
$v + x ((d v)/(d x))$
Check your answer
$v ((d v)/(d x))$ — The product rule on $y=v x$ gives $d y/d x = v + x ((d v)/(d x))$, the plain $v$ term is not multiplied by $d v/d x$, it stands alone.
$(v) + ((d v)/(d x))$ — The product rule gives $x$ multiplying $d v/d x$, not $d v/d x$ alone, dropping that factor loses the $x$ dependence entirely.
$x ((d v)/(d x))$ — Differentiating a product $v x$ gives TWO terms, $v$ and $x ((d v)/(d x))$, dropping the first leaves only half the derivative.
✓ $v + x ((d v)/(d x))$ — (D) Since $y=v x$ is a product of $v(x)$ and $x$, the product rule gives $d y/d x = v + x ((d v)/(d x))$, both terms are needed, with $x$ multiplying $d v/d x$.
Is $d y/d x = (x^2+y^2)/(x y)$ homogeneous?
no, homogeneous equations cannot contain $x^2$ or $y^2$ terms
yes, dividing every term by $x^2$ gives $(1+(y/x)^2)/(y/x)$, a function of $y/x$ alone
yes, but only because both terms happen to be squared
no, the numerator and denominator both have degree 2, so they cancel and no substitution helps
Check your answer
no, homogeneous equations cannot contain $x^2$ or $y^2$ terms — Squared terms are perfectly fine in a homogeneous equation, the test is only whether the whole expression reduces to a function of $y/x$, which this one does.
✓ yes, dividing every term by $x^2$ gives $(1+(y/x)^2)/(y/x)$, a function of $y/x$ alone — (B) Dividing $(x^2+y^2)/(x y)$ through by $x^2$ gives $(1+(y/x)^2)/(y/x)$, every term now written purely in terms of $y/x$, confirming the equation is homogeneous.
yes, but only because both terms happen to be squared — The squares are not themselves the reason, what matters is that dividing every term by $x^2$ leaves an expression written purely in terms of $y/x$.
no, the numerator and denominator both have degree 2, so they cancel and no substitution helps — The numerator and denominator sharing the same degree is exactly why dividing by $x^2$ works, revealing a function of $y/x$ alone, nothing here cancels away or blocks the substitution.
✕MISCONCEPTION
Step 1
y = vx
Step 2
v + x dv/dx = x + vx
Step 3
x dv/dx = x + vx − v
The working for dy/dx = x + y, in three steps. Which step is wrong? Keep your answer.
Here is a working for $d y/d x = x + y$. Which step is wrong?
Find the wrong step
The working for $d y/d x = x + y$
$y = v x$, so $d y/d x = v + x d v/d x$ Put $y = v x$ and use the product rule for $d y/d x$.
$v + x d v/d x = x + v x$ Put both into the equation.
$x d v/d x = x + v x - v$ Take $v$ to the right side, and now separate $v$ from $x$.
Keep your answer, then test the equation with $\lambda$. Here $F(\lambda x, \lambda y) = \lambda x + \lambda y = \lambda (x + y)$, which is not $F(x, y)$. So the equation is not homogeneous. Putting $y = v x$ is allowed, but only a homogeneous equation is sure to separate after it, so expecting this one to separate is the wrong step. The last line cannot be separated, because $x + v x - v$ is not a function of $x$ times a function of $v$. The equation is linear, and a later section solves it with an integrating factor.
dy/dx = 1 + y/x
dy/dx = x + y
Along y = x. Left: 1 + y/x gives slope 2 at every point. Right: x + y gives 1, 2, 3: not homogeneous.
NOT ALLOWED
the x side
dxx + vx − v
the v side
dvx + vx − v
x + vx − v is not a function of x times a function of v. It belongs to neither side alone, so nothing separates.
the v side
dv1/(g(v) − v)
the x side
dx1/x
When homogeneous, x dv/dx = g(v) − v always splits, dv over g(v) − v on one side, dx over x on the other.
For a homogeneous equation the substitution always works. Here is why.
Why it works
Any homogeneous equation separates
$d y/d x = g(y/x)$ A homogeneous equation depends on $x$ and $y$ only through $y/x$.
$v + x d v/d x = g(v)$ Put $y = v x$, so $y/x = v$ and $d y/d x = v + x d v/d x$.
$x d v/d x = g(v) - v$ Take $v$ to the right side.
*$(\mathrm{d} v)/(g(v) - v) = (\mathrm{d} x)/x$* Multiply by $\mathrm{d} x$, then divide by $x$ and by $g(v) - v$, which needs $g(v) \neq v$.
One case is left. Where $g(v) = v$, the equation is $d y/d x = y/x$, which separates already.
The finished form and the unfinished one. At v equals 2 they give minus 5 and minus 1, a gap of 4.
Still, finish the algebra before you separate. Here it is for $d y/d x = (x + y)/(x - y)$. The figure draws the finished form and the unfinished one.
Worked example
Finish the algebra for $d y/d x = (x + y)/(x - y)$
$v + x d v/d x = (x + v x)/(x - v x)$ Put $y = v x$ on both sides.
$v + x d v/d x = (1 + v)/(1 - v)$ Cancel the $x$ above and below the line.
$x d v/d x = (1 + v)/(1 - v) - v$ Take $v$ to the right side.
$x d v/d x = (1 + v - v (1 - v))/(1 - v)$ Write $v$ over the same denominator, $1 - v$.
*$x d v/d x = (1 + v^2)/(1 - v)$* Open the bracket: the top is $1 + v - v + v^2 = 1 + v^2$, not 1.
Check at $v = 2$: $(1 + 4)/(1 - 2) = -5$, but $1/(1 - 2) = -1$ The two forms differ, as the figure shows.
What to do instead
So test with $\lambda$ before you put $y = v x$. Then finish the algebra, and separate.
Check yourself
After $y = v x$, the homogeneous equation $d y/d x = g(y/x)$ becomes $x d v/d x = g(v) - v$. What is true of this equation?
it separates only when $g$ is a polynomial, since other $g$ give a term you cannot separate
it separates only after the right side is simplified
its variables always separate: $d v/(g(v) - v) = d x/x$, where $g(v) \neq v$
it is linear in $v$, so it needs an integrating factor
Check your answer
it separates only when $g$ is a polynomial, since other $g$ give a term you cannot separate — Nothing about $g$ matters but $g(v) \neq v$. The right side is a function of $v$ alone, so dividing by it and by $x$ separates the variables for any $g$.
it separates only after the right side is simplified — Simplifying $g(v) - v$ can make the integral easier, but the variables separate before it: divide both sides by $x (g(v) - v)$.
✓ its variables always separate: $d v/(g(v) - v) = d x/x$, where $g(v) \neq v$ — (C) The right side $g(v) - v$ is a function of $v$ alone, so dividing both sides by $x (g(v) - v)$ gives $d v/(g(v) - v) = d x/x$. A homogeneous equation always separates after $y = v x$. The error to watch for is putting $y = v x$ into an equation that is not homogeneous.
it is linear in $v$, so it needs an integrating factor — It is linear in $v$ only when $g$ is, and it needs no integrating factor: dividing by $x (g(v) - v)$ separates it as it stands.
For $d y/d x=(x+y)/(x-y)$, after substituting $y=v x$ the equation becomes $x ((d v)/(d x)) = (1+v)/(1-v) - v$. What does the right side simplify to?
$(1-v^2)/(1-v)$
$1/(1-v)$
$(1+v)/(1-2v)$
$(1+v^2)/(1-v)$
Check your answer
$(1-v^2)/(1-v)$ — $(1+v)/(1-v)-v = [(1+v)-v(1-v)]/(1-v) = (1+v-v+v^2)/(1-v) = (1+v^2)/(1-v)$, the $v$ and $-v$ cancel, leaving $+v^2$, not $-v^2$.
$1/(1-v)$ — Combining $(1+v)/(1-v)-v$ over the common denominator gives $(1+v^2)/(1-v)$, not $1/(1-v)$, this shortcut drops the $v^2$ term produced by $v(1-v)$.
$(1+v)/(1-2v)$ — The common denominator stays $(1-v)$ throughout the combination, there is no step that doubles $v$ inside the denominator.
At $v=2$, the correctly combined form $(1+v^2)/(1-v)$ and the wrongly-simplified form $1/(1-v)$ give which two values?
$-5$ and $-1$
$5$ and $1$
$-5$ and $-5$
$-1$ and $-5$
Check your answer
✓ $-5$ and $-1$ — (A) At $v=2$: the correct form gives $(1+4)/(1-2)=-5$; the wrongly-simplified form gives $1/(1-2)=-1$, two different numbers, exposing the dropped $v^2$ term.
$5$ and $1$ — At $v=2$, the denominator $1-v=-1$ is negative in both expressions, so both results carry a negative sign, $-5$ and $-1$, not their positive counterparts.
$-5$ and $-5$ — The whole point of checking $v=2$ is that the two forms give DIFFERENT values, $-5$ against $-1$ , they do not agree, which is what exposes the wrong simplification.
$-1$ and $-5$ — $-5$ comes from the correctly combined form; $-1$ comes from the wrongly-simplified one, swapping them attributes each value to the wrong expression.
Left: y = x ln x + Cx for C = −0.4, 0, 0.4. Right: v = y/x = ln x + C, shifted copies.
Four members of y equals x ln x plus C x. The dot at (1, 0) picks the member C equals 0.
Worked example
Solve $d y/d x = (x+y)/x$ by substitution
$d y/d x = (x+y)/x$ Test it: $F(\lambda x, \lambda y) = (\lambda x + \lambda y)/(\lambda x) = (x + y)/x$. So the equation is homogeneous.
$y = v x$, $d y/d x = v + x \cdot d v/d x$ Put $y = v x$. By the product rule, $d y/d x = v + x d v/d x$.
$v + x \cdot d v/d x = 1 + v$ Put both into the equation. The right side is $1 + y/x = 1 + v$.
$x \cdot d v/d x = 1$ The $v$ on each side cancels. What is left separates.
$\mathrm{d} v = (\mathrm{d} x)/x$ Separate the variables. Here $x \neq 0$.
$v = \ln(x) + C$ Integrate both sides. Take $x > 0$, so $\ln \left|x\right|$ is $\ln x$.
$y = x \cdot \ln(x) + C x$ Put $v = y/x$ back, and multiply both sides by $x$.
*$d y/d x = \ln(x) + 1 + C = (x+y)/x$* Check the answer. Its derivative is $\ln x + 1 + C$, which equals $(x + y)/x$.
Check yourself
In solving $d y/d x=(x+y)/x$ via $y=v x$, why does the $v$ term cancel from both sides?
because the right side simplifies to $1+v$, matching the $v$ already on the left
because the equation had only one arbitrary constant to begin with
because $v$ always cancels whenever a homogeneous equation is substituted, as a rule
because $x$ was divided out of the equation before substituting
Check your answer
✓ because the right side simplifies to $1+v$, matching the $v$ already on the left — (A) Substituting $y=v x$ into the right side gives $(x+v x)/x=1+v$, exactly matching the $v$ the product rule already placed on the left, the two cancel because of THIS equation’s specific form, not by any general rule.
because the equation had only one arbitrary constant to begin with — Constant count is a fact about a FAMILY of curves used when forming an equation, it has nothing to do with why a term cancels during solving.
because $v$ always cancels whenever a homogeneous equation is substituted, as a rule — There is no universal rule that $v$ always cancels, it happens here because $(x+y)/x$ simplifies to exactly $1+v$, matching the $v$ already present on the left. A different right side would not cancel the same way.
because $x$ was divided out of the equation before substituting — No separate division step happens here, substituting $y=v x$ into $(x+y)/x$ directly gives $(x+v x)/x=1+v$, which is what matches the $v$ on the left.
Solve $d y/d x=(x+2 y)/x$ using $y=v x$. What is the general solution?
$y = K x^2 - x$
$y = x \ln(x) + K x$
$y = K x - 1$
$y = x^2 + K x$
Check your answer
✓ $y = K x^2 - x$ — (A) Substituting $y=v x$ gives $x ((d v)/(d x))=1+v$, which separates to $d v/(1+v)=d x/x$; integrating gives $1+v=K x$, so $v=K x-1$, and multiplying back by $x$ gives $y=K x^2-x$.
$y = x \ln(x) + K x$ — $y=x \ln(x)+K x$ solves the original equation, $d y/d x=(x+y)/x$. The coefficient $2$ here changes the equation enough that the solution family is $y=K x^2-x$ instead, a different right side needs its own solving through, not the same answer shape.
$y = K x - 1$ — $v=K x-1$ is correct as far as it goes, but $v$ is only the auxiliary variable, multiplying by $x$ (since $y=v x$) gives $y=K x^2-x$, the answer in the original variables.
$y = x^2 + K x$ — The arbitrary constant $K$ belongs with the $x^2$ term, since $v=K x-1$ multiplied by $x$ gives $y=K x^2-x$, the $-x$ term carries no free constant at all.
Once $v=\ln(x)+C$ is found for $d y/d x=(x+y)/x$, why is substituting $v=y/x$ back the final step, not a further step of solving?
because $v$ was only ever a stand-in, not the equation’s real unknown
because the equation would otherwise stop being homogeneous
because $v$ cannot be integrated any further once it reaches this form
because it converts the equation from order 1 to order 0
Check your answer
✓ because $v$ was only ever a stand-in, not the equation’s real unknown — (A) $v$ was introduced purely as a convenience so the equation would separate, once solved for $v$, substituting $v=y/x$ back is simply returning the answer to the variables the original question was asked in.
because the equation would otherwise stop being homogeneous — Homogeneity describes the original equation before substitution; it is not a property that solving the $v$-equation could add or remove.
because $v$ cannot be integrated any further once it reaches this form — Integration is already finished by the time $v=\ln(x)+C$ is reached, substituting $v=y/x$ back is about returning to the original variables, not about hitting a limit on integration.
because it converts the equation from order 1 to order 0 — Order is a property of a differential equation, not of an already-solved expression, by the time $v=\ln(x)+C$ is written down, solving is done, and no order applies to it any more.
A first-order equation is LINEAR when it has the form
$d y/d x + P y = Q$,
where $P$ and $Q$ are functions of $x$ alone, or constants. So $y$ and $d y/d x$ appear only to the first power, and never multiply each other.
Multiply both sides by the INTEGRATING FACTOR, $\text{IF} = e^{\int P \mathrm{d} x}$.
Why it works
When the integral is a logarithm
$\int P \mathrm{d} x = \ln \left|u\right|$ Suppose the integral works out to the log of the absolute value of some function $u$ of $x$.
$\text{IF} = e^{\ln \left|u\right|} = \left|u\right|$ Put this into $\text{IF} = e^{\int P \mathrm{d} x}$: the exponential undoes the log.
$\text{IF} = u$ On an interval where $u$ keeps one sign, $\left|u\right|$ and $u$ agree, so $u$ itself serves as $\text{IF}$.
$\int 1/x \mathrm{d} x = \ln \left|x\right|$, so $\text{IF} = x$ for $x > 0$ Example: here $u = x$, which keeps one sign on $x > 0$.
Then the left side becomes one derivative. Here is why, line by line:
Why it works
The factor makes one derivative
$\text{IF} d y/d x + P y \text{IF} = Q \text{IF}$ Multiply both sides of $d y/d x + P y = Q$ by the integrating factor $\text{IF}$.
$d/d x (\text{IF}) = P \text{IF}$ By the chain rule: the power of $e$ has derivative $P$.
$\text{IF} d y/d x + y d/d x (\text{IF}) = Q \text{IF}$ Write $P \text{IF}$ as the derivative of $\text{IF}$.
*$d/d x (y \cdot \text{IF}) = Q \cdot \text{IF}$* Read the product rule backwards: the left side is the derivative of $y \cdot \text{IF}$.
Check: $d/d x (y \cdot \text{IF}) = \text{IF} d y/d x + y P \text{IF}$ Differentiate $y \cdot \text{IF}$ forwards by the product rule to get the left side of line 1.
top=0.75right=1
y × IF as a 2 by 3 rectangle. Growing it adds IF dy = 0.75, y d(IF) = 1 and a tiny corner.
Slope marks of dy/dx + y = 2, with y = 2 + Ce⁻ˣ for C = −2 to 2. Every member nears y = 2.
For y = 2 + e⁻ˣ, the product y eˣ is 2eˣ + 1, top. Its derivative, below, is 2eˣ: Q times the factor.
Let us run it on $d y/d x + y = 2$. Here $P = 1$ and $Q = 2$.
Worked example
Solve $d y/d x + y = 2$
$\text{IF} = e^{\int 1 \mathrm{d} x} = e^x$ Find the factor from $P = 1$.
$e^x d y/d x + e^x y = 2 e^x$ Multiply both sides by $e^x$.
$d/d x (y e^x) = 2 e^x$ The left side is the product rule read backwards.
$y e^x = 2 e^x + C$ Integrate both sides, with one $+ C$.
*$y = 2 + C e^{-x}$* Divide both sides by $e^x$.
Check: $d y/d x + y = -C e^{-x} + 2 + C e^{-x} = 2$ Put the answer back into the equation.
Any integral of $P$ gives a working factor: a $+ C$ inside it would multiply the factor by $e^C$, which cancels from both sides. So leave out a $+ C$ inside the factor.
When x depends on y
With $x$ and $y$ swapped, an equation $d x/d y + P_1 x = Q_1$ has $P_1$ and $Q_1$ as functions of $y$ alone. Its factor is $e^{\int P_1 \mathrm{d} y}$, and $x \cdot \text{IF} = \int Q_1 \cdot \text{IF} \mathrm{d} y + C$.
Your turn: which of these is linear: $d y/d x + x y = x$, or $d y/d x + y^2 = x$? For the linear one, name $P$ and $Q$.
Check yourself
A first-order ODE is LINEAR when it fits the standard form \_\_\_\_\_\_.
$d y/d x = P(x) y + Q(x) y^2$
$y ((d y)/(d x)) + P(x) = Q(x)$
$d y/d x + P(x) y^2 = Q(x)$
$d y/d x + P(x) y = Q(x)$
Check your answer
$d y/d x = P(x) y + Q(x) y^2$ — A $y^2$ term breaks linearity, the standard linear form allows only $y$ to the first power, alongside its derivative.
$y ((d y)/(d x)) + P(x) = Q(x)$ — Multiplying $y$ by $d y/d x$ is a non-linear combination, the standard linear form keeps $d y/d x$ and $y$ as separate, individually first-power terms.
$d y/d x + P(x) y^2 = Q(x)$ — $y^2$ in place of $y$ breaks linearity, the standard form needs $y$ to the first power exactly.
✓ $d y/d x + P(x) y = Q(x)$ — (D) $d y/d x + P(x) y = Q(x)$ keeps both $d y/d x$ and $y$ to the first power, with no product between them, any squared $y$ term or direct multiplication of $y$ by its derivative breaks this linear form.
What is the integrating factor for $d y/d x + P(x) y = Q(x)$?
$e^{P(x)}$
$\int P d x$
$e^{\int P d x}$
$e^{\int Q d x}$
Check your answer
$e^{P(x)}$ — $P(x)$ must be integrated first, exponentiating it directly, with no integral, skips the step that makes the integrating factor work.
$\int P d x$ — The integrating factor is $e$ RAISED to the integral of $P$, not the integral by itself, forgetting the exponential leaves an unrelated quantity.
✓ $e^{\int P d x}$ — (C) The integrating factor is $\text{IF} = e^{\int P d x}$, built from $P(x)$ alone, integrated first and then placed as the exponent of $e$.
$e^{\int Q d x}$ — The integrating factor is built from $P(x)$, the coefficient multiplying $y$, $Q(x)$ plays no role in building it at all.
Multiplying through by the integrating factor turns the left side of the equation into \_\_\_\_\_\_.
derivative $d/d x (y \cdot \text{IF})$
a constant, independent of $x$
a separable expression in $y$ alone
the second derivative of $y$
Check your answer
✓ derivative $d/d x (y \cdot \text{IF})$ — (A) The integrating factor is chosen precisely so that $d y/d x + P(x) y$ collapses into $d/d x (y \cdot \text{IF})$, the derivative of one product, not a separated form, a higher derivative, or a constant.
a constant, independent of $x$ — The left side stays a function of $x$, now written as $d/d x (y \cdot \text{IF})$, nothing about multiplying by the integrating factor turns it into a fixed constant.
a separable expression in $y$ alone — Separating in $y$ alone is exactly what a linear equation resists doing directly , the integrating factor is a different tool, turning the left side into one derivative instead.
the second derivative of $y$ — The result is still a first derivative, just of the PRODUCT $y \cdot \text{IF}$, no second derivative appears anywhere in the method.
What is the integrating factor for $d y/d x + 2 y = x$?
$e^{2 x}$
$2 e^{x}$
$e^{2 x^2}$
$e^{x}$
Check your answer
✓ $e^{2 x}$ — (A) $P(x)=2$, a constant, so $\int P d x = 2 x$, giving the integrating factor $e^{2 x}$.
$2 e^{x}$ — The $2$ belongs INSIDE the exponent, as part of $\int 2 d x = 2 x$, multiplying the whole exponential by $2$ afterward is a different, incorrect operation.
$e^{2 x^2}$ — Integrating the CONSTANT $2$ with respect to $x$ gives $2 x$, not $2 x^2$, that extra power would come from integrating $2 x$, not the constant $2$.
$e^{x}$ — $P(x)=2$ here, so $\int P d x = 2 x$, giving $e^{2 x}$, dropping the coefficient $2$ leaves the wrong exponent.
For y = x²/3 + 1/x, the product y × x is x³/3 + 1, top. Its derivative, below, is x², Q × IF.
Four members of y equals x squared over 3 plus C over x. They start apart and close in, as C over x fades.
Worked example
Solve $d y/d x + (1/x) y = x$ using the integrating factor
$d y/d x + (1/x) y = x$, $x > 0$ This has the form $d y/d x + P y = Q$, so it is linear. Here $x > 0$.
$P(x) = 1/x$, $Q(x) = x$ Read off $P$ and $Q$. Both are functions of $x$ alone.
$y \cdot x = \int x \cdot x \mathrm{d} x + C$ Use $y \cdot \text{IF} = \int Q \cdot \text{IF} \mathrm{d} x + C$.
$y \cdot x = x^3/3 + C$ Integrate $x^2$.
$y = x^2/3 + C/x$ Divide both sides by $x$.
*$d y/d x + y/x$ simplifies back to $x$* Check the answer. $d y/d x = 2x/3 - C/x^2$, and adding $y/x = x/3 + C/x^2$ gives $x$.
Check yourself
For $d y/d x + (1/x) y = x$, why is $\text{IF} = e^{\int (1/x) d x}$ equal to just $x$?
because the exponential and the integral cancel for any $P(x)$
because $\int (1/x) d x = \ln(x)$, and $e^{\ln(x)} = x$
because integrating $1/x$ gives $x$ directly
because $P(x)=1/x$ already equals $x$ when inverted
Check your answer
because the exponential and the integral cancel for any $P(x)$ — $e^{\int P d x}$ simplifying to something as clean as $x$ is specific to $P(x)=1/x$ here, for most choices of $P(x)$, the integrating factor stays a more complicated exponential expression.
✓ because $\int (1/x) d x = \ln(x)$, and $e^{\ln(x)} = x$ — (B) $\int (1/x) d x = \ln(x)$, and $e^{\ln(x)}$ simplifies to $x$ by definition of the natural log, two clean steps, not a shortcut that works for every $P(x)$.
because integrating $1/x$ gives $x$ directly — $\int (1/x) d x = \ln(x)$, not $x$. The integrating factor reaches $x$ only in the NEXT step, once $e^{\ln(x)}$ is simplified.
because $P(x)=1/x$ already equals $x$ when inverted — Inverting $P(x)$ is not part of the method, the integrating factor comes from integrating $P(x)$ first, then raising $e$ to that result, which happens to simplify to $x$ here.
Once $y \cdot x = \int x \cdot x d x + C$ is reached for $d y/d x+(1/x) y=x$, what is the general solution?
$y = x^2/3 + C/x$
$y = x^2/3 + C$
$y = x^3/3 + C$
$y = (x^2) + C/x$
Check your answer
✓ $y = x^2/3 + C/x$ — (A) $\int x \cdot x d x = x^3/3$, so $y \cdot x=x^3/3+C$; dividing every term by $x$ gives $y=x^2/3+C/x$.
$y = x^2/3 + C$ — Both terms on the right of $y \cdot x=x^3/3+C$ need dividing by $x$, the constant term becomes $C/x$, not a bare $C$.
$y = x^3/3 + C$ — $y \cdot x = x^3/3+C$ still needs dividing through by $x$ to isolate $y$, skipping that step leaves $y$ mixed with an extra factor of $x$.
$y = (x^2) + C/x$ — Integrating $x \cdot x=x^2$ gives $x^3/3$, and dividing by $x$ gives $x^2/3$, the $/3$ from integrating a power is not optional.
Checking the solution $y = x^2/3 + C/x$ against the original equation confirms $d y/d x + y/x$ simplifies back to \_\_\_\_\_\_.
$0$
$x$
$2 x/3$
$x^2/3$
Check your answer
$0$ — Nothing here cancels to zero, adding $d y/d x$ and $y/x$ together reproduces the original right side, $x$, term by term.
✓ $x$ — (B) $d y/d x=2 x/3-C/x^2$, and $y/x=x/3+C/x^2$; adding the two, the $C/x^2$ terms cancel and $2x/3+x/3=x$, exactly the original right side.
$2 x/3$ — $d y/d x=2 x/3-C/x^2$ is only half the check, adding $y/x$ as well is what makes the two terms combine back to $x$.
$x^2/3$ — $x^2/3$ is part of the SOLUTION, not the result of the check, substituting back and simplifying must reproduce the original right side, $x$.
1,000 rupees at 5 per cent, added at every moment. It gains 50 a year at the start, 82.44 a year after 10 years.
Go back to the bank account. Say the bank pays 5 per cent a year, added at every moment. Then at each moment your money grows at a rate of 0.05 times the amount there. Double the money, and it grows twice as fast.
A town at the edge of its fields. New houses are going up where the old town ends.
A town of 50,000 people, growing continuously at 2 per cent a year, reaches about 74,591 in 20 years.
A town’s population behaves the same way. You have met Census figures in Social Science. More people means more births. So while birth and death rates stay steady, the population grows at a rate set by its size.
Rate against amount. Bank: 50 a year on 1,000, 100 on 2,000. Tea: a 60-degree gap shrinks 6 a minute, 30 shrinks 3.
The equation they share
In both cases the rate is a fixed number times the amount. Write $y$ for the amount at time $t$. Then
$d y/d t = k y$.
Slope marks of dy/dt = 0.4y: the higher the point, the steeper the slope mark. Members: y₀ times e to the 0.4t, y₀ = 0.5 to 2.
The gap between tea and room starts at 60 degrees, k = −0.1: 22.1 at 10 minutes, 8.1 at 20. The tea nears 30 °C.
The sign of k
Here $k > 0$ for growth, and $k < 0$ for decay. The tea fits too, if $y$ is how much hotter the tea is than the room. That gap shrinks, so its $k$ is negative.
Separating gives ln y = kt + C. Left, y = 1.5 times e to the 0.4t. Right, ln y: a line of slope 0.4.
Let us solve it. The equation separates, and the amount $y$ is positive.
Worked example
Solve $d y/d t = k y$
$\mathrm{d} y = k y \mathrm{d} t$ Multiply both sides by $\mathrm{d} t$.
$(\mathrm{d} y)/y = k \mathrm{d} t$ Divide both sides by $y$, which is positive, so $y \neq 0$.
$\ln y = k t + C$ Integrate both sides with one $+ C$, and $\ln \left|y\right| = \ln y$ as $y > 0$.
$y = e^{k t + C} = e^C e^{k t}$ Exponentiate both sides, since $e^{\ln y} = y$, then split the power.
*$y = y_0 e^{k t}$* The constant $e^C$ is positive and is $y$ at $t = 0$, so call it $y_0$, the starting amount.
Check: $d y/d t = k y_0 e^{k t} = k y$ Differentiate the answer: it gives back the equation.
So every solution has the form $y = y_0 e^{k t}$, where $y_0$ is the amount at $t = 0$.
Both start at 1.5. With k equal to 0.4 the curve climbs to 4.98 by t equals 3. With minus 0.4 it falls to 0.45.
The figure starts two curves at $y_0 = 1.5$. With $k = 0.4$ the amount climbs to about 4.98 by $t = 3$. With $k = -0.4$ it falls to about 0.45. It gets close to 0 but never reaches it.
Doubling the money
Your turn: money grows at 5 per cent a year, added at every moment. Is $k$ equal to 5 or to 0.05? About how many years does the money take to double?
Check: $k = 0.05$. The money doubles when $e^{0.05 t} = 2$, so $t = \ln(2)/0.05 \approx 13.9$ years.
A SCHOLAR INDIA REMEMBERS
Predict the tea with one steady rate. It is at $90$ °C in a room at $30$ °C, and at first it cools $6$ degrees a minute. At that rate it reaches $30$ °C in $10$ minutes, then goes colder than the room. No cup of tea does that. So change the model, not the reading. The rate is set by the gap, $d y/d t = -0.1 y$. After $10$ minutes the tea is at about $52.1$ °C.
Brahmagupta · the brahminy kite
Check yourself
The model $d y/d t = k y$ describes quantities that change at a rate proportional to \_\_\_\_\_\_.
the time elapsed since $t=0$
a fixed constant rate, independent of the value itself
the square of their current value
their own current value
Check your answer
the time elapsed since $t=0$ — A rate depending only on elapsed time, not on $y$ itself, would be $d y/d t=k$, this chapter’s model instead has the rate scale with $y$’s own current size.
a fixed constant rate, independent of the value itself — A fixed, value-independent rate is the opposite of this model, here, the larger $y$ already is, the faster it changes, since the rate is $k$ TIMES $y$.
the square of their current value — $d y/d t=k y$ has the rate proportional to $y$ itself, to the first power, a $y^2$ dependence would be a different, non-linear model entirely.
✓ their own current value — (D) $d y/d t = k y$ makes the rate of change proportional to $y$’s own current value, not to elapsed time, not to $y^2$, and not to a fixed, unchanging amount.
In $d y/d t = k y$, what does the sign of $k$ determine?
growth ($k>0$) versus decay ($k<0$)
whether the equation is separable
the order of the differential equation
the initial value $y_0$
Check your answer
✓ growth ($k>0$) versus decay ($k<0$) — (A) The sign of $k$ alone decides direction: positive $k$ makes $y$ grow over time, negative $k$ makes it decay, everything else about the equation stays unaffected by that sign.
whether the equation is separable — $d y/y=k d t$ separates identically whichever sign $k$ carries, the sign only affects the DIRECTION of change, not whether the method applies.
the order of the differential equation — Order comes from the highest derivative present, always first order here, the sign of $k$ has nothing to do with it.
the initial value $y_0$ — $y_0$ comes from the condition at $t=0$, entirely independent of $k$, the sign of $k$ only decides whether $y$ increases or decreases from there.
Solving $d y/d t = k y$ by separating variables gives $\ln(y) = k t + C$. What is the general solution for $y$?
$y = y_0 e^{k t}$
$y = k t + y_0$
$y = e^{k t}$
$y = e^{k t} + y_0$
Check your answer
✓ $y = y_0 e^{k t}$ — (A) Exponentiating $\ln(y)=k t+C$ gives $y=e^C e^{k t}$; renaming the constant $e^C$ as $y_0$, the value at $t=0$, gives $y=y_0 e^{k t}$.
$y = k t + y_0$ — $y = k t + y_0$ is a straight line, which a CONSTANT rate $d y/d t = k$ gives. Here the rate depends on $y$, so exponentiate $\ln(y) = k t + C$ instead of dropping the $\ln$.
$y = e^{k t}$ — Exponentiating $\ln(y) = k t + C$ gives $y = e^C e^{k t}$. Dropping $e^C$ leaves only the solution that starts at $y = 1$; the constant, renamed $y_0$, multiplies $e^{k t}$.
$y = e^{k t} + y_0$ — Exponentiating $\ln(y)=k t+C$ gives $y=e^C \cdot e^{k t}$, a PRODUCT with the constant $e^C$ renamed $y_0$ , not a sum.
Start with 4 grams, and half is gone in 20 years. So 2 grams are left at 20 years, and 1 gram at 40.
Worked example
Half-life: find the decay constant from “50% in 20 years”
$d y/d t = k \cdot y$ The rate is proportional to the amount. For decay, $k < 0$.
$(\mathrm{d} y)/y = k \mathrm{d} t$ Separate the variables. The amount $y$ is positive, so you can divide by it.
$\ln(y) = k t + C$ Integrate both sides. Since $y > 0$, $\ln \left|y\right|$ is $\ln y$.
$y = y_0 e^{k t}$ Exponentiate both sides, and call $e^C$ by what it is, $y_0$, the amount at $t = 0$.
$y_0/2 = y_0 e^{20 \cdot k}$ Half is left after 20 years. So put $t = 20$ and $y = y_0/2$.
$k = -\ln(2)/20$ Cancel $y_0$, so $e^{20 k} = 1/2$. Take logs: $k = -\ln(2)/20$, negative, as decay needs.
*$y = y_0 \cdot 2^{-t/20}$* Put $k$ back. Since $e^{-\ln(2) t/20} = 2^{-t/20}$, this is the amount left after $t$ years.
Check: $t = 0$ gives $y_0 \cdot 2^0 = y_0$, and $t = 20$ gives $y_0 \cdot 2^{-1} = y_0/2$ Put both given facts back. The amount starts at $y_0$ and halves by year 20. The answer also fits the equation, since $y_0 e^{k t}$ has derivative $k y$.
A SCHOLAR INDIA REMEMBERS
The constant came out negative, $k = -\ln(2)/20$, about $-0.0347$. Is a negative number allowed here? As a rate, yes. A negative rate is an amount going down, and decay needs $k < 0$. As an amount, no. So check that $y$ never goes below $0$. Start with $4$ grams. Then $2$ are left at $20$ years, $1$ at $40$ and $0.5$ at $60$. Each step halves, so the amount nears $0$ and never reaches it.
Brahmagupta · the brahminy kite
Check yourself
With this chapter’s model $d y/d t = k y$ ($k < 0$ for decay), a radioactive substance decays at a rate proportional to the amount present, and half of it decays in 20 years. What equation does this give for $k$?
$e^{20 k} = 2$
$20 k = 1/2$
$e^{20 k} = 1/2$
$e^{-20 k} = 1/2$
Check your answer
$e^{20 k} = 2$ — Half remaining means the ratio $y/y_0$ is $1/2$, not $2$. Inverting the fraction reverses what “half decays” means.
$20 k = 1/2$ — The relationship between $y$ and $t$ is exponential, $y = y_0 e^{k t}$. Dropping the exponential leaves an equation with no basis in the decay model.
✓ $e^{20 k} = 1/2$ — (C) At $t = 20$, half remains: $y_0/2 = y_0 e^{20 k}$. Cancelling $y_0$ gives $e^{20 k} = 1/2$, the equation $k$ must satisfy, and it makes $k < 0$, as decay needs.
$e^{-20 k} = 1/2$ — $e^{-20 k} = 1/2$ belongs to the model $y = y_0 e^{-k t}$ with a positive decay constant, which this chapter does not use. In $d y/d t = k y$ the minus sign is inside $k < 0$, so the equation is $e^{20 k} = 1/2$.
With this chapter’s model $d y/d t = k y$ ($k < 0$ for decay), if instead half the substance decays in 10 years, what is $k$?
$k = -\ln(2)/20$
$k = -\ln(2)/10$
$k = -10 \ln(2)$
$k = \ln(2)/10$
Check your answer
$k = -\ln(2)/20$ — $-\ln(2)/20$ solves the original 20-year version. With a 10-year half-life the equation is $e^{10 k} = 1/2$, giving $k = -\ln(2)/10$.
✓ $k = -\ln(2)/10$ — (B) With a 10-year half-life, $e^{10 k} = 1/2$, so $10 k = \ln(1/2) = -\ln(2)$, giving $k = -\ln(2)/10$. It is negative, as decay needs.
$k = -10 \ln(2)$ — Solving $e^{10 k} = 1/2$ for $k$ divides $\ln(1/2)$ by the half-life, $10$. It does not multiply by it: $k = -\ln(2)/10$.
$k = \ln(2)/10$ — $\ln(2)/10$ is positive, so $e^{k t}$ would grow. In $d y/d t = k y$ decay needs $k < 0$: from $10 k = \ln(1/2) = -\ln(2)$, $k = -\ln(2)/10$.
With this chapter’s model $d y/d t = k y$ ($k < 0$ for decay), once $k = -\ln(2)/20$ the remaining amount $y = y_0 e^{k t}$ can be rewritten as $y = y_0 \cdot 2^{-t/20}$. Why does $e^{k t}$ equal $2^{-t/20}$ here?
because $t/20$ always simplifies to a power of 2, whatever $t$ is
because $e$ and $2$ can always be swapped as the base of any exponential
because $\ln(2)$ is approximately equal to $20$
because $k = -\ln(2)/20$: $e^{k t} = (e^{\ln(2)})^{-t/20} = 2^{-t/20}$
Check your answer
because $t/20$ always simplifies to a power of 2, whatever $t$ is — $t/20$ is simply the exponent applied to the base $2$, it is not itself claimed to be a power of 2, and needs no such property to make the rewrite work.
because $e$ and $2$ can always be swapped as the base of any exponential — $e$ and $2$ are not interchangeable bases in general. The swap works here only because $k$ was built from $-\ln(2)/20$, and $e^{\ln(2)}$ is exactly $2$.
because $\ln(2)$ is approximately equal to $20$ — $\ln(2) \approx 0.693$ and $20$ (years) are two entirely different numbers that happen to appear in the same formula, nothing sets them equal to each other.
✓ because $k = -\ln(2)/20$: $e^{k t} = (e^{\ln(2)})^{-t/20} = 2^{-t/20}$ — (D) Because $k = -\ln(2)/20$, $e^{k t} = e^{-(t/20) \ln(2)} = (e^{\ln(2)})^{-t/20}$, and $e^{\ln(2)} = 2$ by definition. That gives $2^{-t/20}$ exactly, a rewrite that depends entirely on how $k$ was built.
Seven idea cards. Order and degree lead to general and particular solutions, then to forming, separable and linear. Homogeneous and growth build on separable.
The first page asked a question. When a rate depends on the amount itself, how do you find the amount?
separable
y² cos xx y²
homogeneous
(x + y)/(x − y)(x² + y²)/(x y)
linear
x + yx − y/x
none of the three
x + y²
The right side of dy/dx, sorted: separable, homogeneous, linear, or none of the three. Read it before you choose a move.
A DIFFERENTIAL EQUATION holds an unknown function and its derivatives. Its answer is a function.
ORDER is the order of the highest derivative. DEGREE is that derivative’s power, and it exists only when the equation is a polynomial in all its derivatives.
The GENERAL SOLUTION of an equation of order $n$ has $n$ arbitrary constants. One given fact per constant picks a PARTICULAR SOLUTION.
FORMING AN EQUATION (beyond the board syllabus): a family with $n$ constants fits an equation of order $n$. Differentiate $n$ times, then eliminate the constants.
VARIABLES SEPARABLE: $d y/d x = g(x) h(y)$. Where $h(y) \neq 0$, $\int (\mathrm{d} y)/h(y) = \int g(x) \mathrm{d} x + C$.
HOMOGENEOUS: test $F(\lambda x, \lambda y) = F(x, y)$ first. Then put $y = v x$, finish the algebra, and separate.
LINEAR: $d y/d x + P y = Q$, with $P$ and $Q$ functions of $x$ alone. With $\text{IF} = e^{\int P \mathrm{d} x}$, $y \cdot \text{IF} = \int Q \cdot \text{IF} \mathrm{d} x + C$.
GROWTH AND DECAY: $d y/d t = k y$ gives $y = y_0 e^{k t}$. Growth has $k > 0$, and decay has $k < 0$.
Three panels, one per method: separable, homogeneous and linear. Each draws two members of the family that method gives.
Here is the answer. Three types of first-order equation, one move: turn the equation into something you can integrate. There are three ways to make that move: separate the variables, put $y = v x$, or multiply by an integrating factor. Each ends in an integral from Chapter 7. The answer is a family with one constant. One given value picks one member, as the tank’s starting level picked one $+ C$ in Chapter 7.
Money at 5 per cent, added every moment: y = y₀ times e to the 0.05t. Starting at 1,000 picks y₀ = 1000.
So the bank account has its answer. Money at 5 per cent a year, added at every moment, follows $y = y_0 e^{0.05 t}$. Put in 1,000 rupees, and the starting value picks $y_0 = 1000$.
Check yourself
Under “Three types, one move”, the move is to turn an equation into something you can integrate. How does a variables-separable equation make that move, compared with the other two kinds?
it is the only kind that can be solved exactly
it needs no substitution, no integrating factor: separate, integrate
it always gives a simpler answer than a homogeneous or linear equation
it never needs to be integrated at all
Check your answer
it is the only kind that can be solved exactly — Homogeneous and linear equations are solved just as exactly. Each needs one step first, a substitution or an integrating factor, and then it is integrated.
✓ it needs no substitution, no integrating factor: separate, integrate — (B) A separable equation needs no step before the move. You separate the variables, then integrate. A homogeneous equation first needs $y = v x$ to become separable, and a linear one first needs the integrating factor.
it always gives a simpler answer than a homogeneous or linear equation — The comparison is about the steps, not the answer. A separable equation’s answer can be just as involved as a homogeneous or linear one’s.
it never needs to be integrated at all — Integration is still needed to finish. What a separable equation skips is the step before it: no substitution and no integrating factor.
Under “Three types, one move”, the list says of a homogeneous equation: put $y = v x$, finish the algebra, and separate. What does putting $y = v x$ do?
it removes the need to integrate the equation at all
it raises the order of the equation so it can be solved
it converts the equation into the linear standard form instead
it turns the equation into a separable one in $v$ and $x$
Check your answer
it removes the need to integrate the equation at all — Integration still follows. The substitution only brings the equation to a separable form; you separate it and then integrate.
it raises the order of the equation so it can be solved — Putting $y = v x$ changes the variable, not the order. The equation is first order before and after.
it converts the equation into the linear standard form instead — $y = v x$ leads to a separable equation in $v$ and $x$, not to the linear form. The integrating factor belongs to the linear method.
✓ it turns the equation into a separable one in $v$ and $x$ — (D) After $y = v x$ and the algebra, the equation is separable in $v$ and $x$. You separate the variables, then integrate. The order stays one, and no integrating factor is used.
Under “Three types, one move”, the list solves a linear equation with $\text{IF} = e^{\int P d x}$. What does multiplying through by $\text{IF}$ do?
it eliminates the arbitrary constant from the equation before integrating
it substitutes a new variable for $y$, the same way $y=v x$ does for a homogeneous equation
it turns the left side into the derivative of one product, $d/d x (y \cdot \text{IF})$
it separates the $x$-terms and $y$-terms onto opposite sides
Check your answer
it eliminates the arbitrary constant from the equation before integrating — Eliminating constants belongs to forming an equation from a family of curves. Solving a linear equation is a different task.
it substitutes a new variable for $y$, the same way $y=v x$ does for a homogeneous equation — The integrating factor multiplies the whole equation. It brings in no new variable in place of $y$; that is the homogeneous method’s step.
✓ it turns the left side into the derivative of one product, $d/d x (y \cdot \text{IF})$ — (C) Multiplying by $\text{IF}$ turns the left side into $d/d x (y \cdot \text{IF})$, one derivative, so the equation reads $d/d x (y \cdot \text{IF}) = Q \cdot \text{IF}$. You integrate both sides. It is not a substitution, a separation or a way to remove a constant.
it separates the $x$-terms and $y$-terms onto opposite sides — A linear equation usually cannot be separated. The integrating factor does something else: it turns the left side into one derivative, which you then integrate.
The section headed “Three types, one move” names one move behind all three methods. What is it?
always substitute $y=v x$ first, whatever the equation looks like
reduce every equation to the literal variables-separable form, even when it needs no substitution
compute the integrating factor before doing anything else
turn the equation into something you can already integrate
Check your answer
always substitute $y=v x$ first, whatever the equation looks like — $y = v x$ is the homogeneous method’s own substitution. Separable and linear equations reach something they can integrate in other ways.
reduce every equation to the literal variables-separable form, even when it needs no substitution — The move is to reach something you can integrate, not always the separable form. A linear equation reaches $d/d x (y \cdot \text{IF}) = Q \cdot \text{IF}$, which is not separated: you integrate it as it stands.
compute the integrating factor before doing anything else — An integrating factor belongs to the linear method. A separable equation needs none.
✓ turn the equation into something you can already integrate — (D) The one move is to turn the equation into something you can integrate. A separable equation is separated, then integrated. A homogeneous one becomes separable after $y = v x$. A linear one becomes $d/d x (y \cdot \text{IF}) = Q \cdot \text{IF}$, which you integrate as it stands.
Check yourself: the whole chapter
For $d y/d x = 3 x^2$, one valid answer is $y = x^3 + C$ for ANY constant $C$. What kind of mathematical object is “the solution” here?
A single curve: the one you get by taking $C = 0$.
A single number: the value $y$ works out to.
A family of curves: one curve for each choice of $C$.
Not yet a solution: a specific point is still needed to fix $C$.
Check your answer
A single curve: the one you get by taking $C = 0$. — No value of $C$ is singled out. $C = 0$ gives one member of the family, but $C = 5$, $C = -2$, and every other value work just as well.
A single number: the value $y$ works out to. — Solving here means finding a FUNCTION, not a number. $y = x^3 + C$ names a whole set of curves, all satisfying the equation.
✓ A family of curves: one curve for each choice of $C$. — (C) Substituting $y = x^3 + C$ back gives $d y/d x = 3 x^2$ for every value of $C$, so the equation is satisfied by a whole family of curves, not one curve or one number.
Not yet a solution: a specific point is still needed to fix $C$. — $y = x^3 + C$ already satisfies the equation for every $C$, with no condition needed. It is a complete answer, a family, not a single curve.
For $((d^2 y)/(d x^2))^2 - ((d y)/(d x))^4 = 0$, what are its order and degree?
Order 2, degree 2: both read from $(d^2 y)/(d x^2)$.
Order 2, degree 4: read from the first-derivative term instead.
Order 4, degree 2: order and degree the wrong way round.
Degree undefined: a squared derivative breaks the polynomial rule.
Check your answer
✓ Order 2, degree 2: both read from $(d^2 y)/(d x^2)$. — (A) The highest derivative present is $(d^2 y)/(d x^2)$, fixing order at 2. It is raised to the power 2, fixing degree at 2 as well.
Order 2, degree 4: read from the first-derivative term instead. — Degree comes only from the HIGHEST-order derivative present. That is $(d^2 y)/(d x^2)$, raised to the power 2, the first derivative’s power does not count.
Order 4, degree 2: order and degree the wrong way round. — Order is which derivative is highest: 2, from $(d^2 y)/(d x^2)$. Degree is the power on that same term: 2. Neither count is 4.
Degree undefined: a squared derivative breaks the polynomial rule. — A whole-number power on a derivative, like squaring it, is exactly what a polynomial term looks like. Degree is undefined only when a derivative sits inside something like $\sin$ or $\ln$.
Every differential equation has a well-defined ORDER. Does every differential equation also have a well-defined DEGREE?
Yes. Degree is defined exactly when order is.
No. Degree needs the equation to be a polynomial in its derivatives.
No. Degree is undefined whenever more than one derivative term is present.
Yes. But only after the equation is solved for $y$.
Check your answer
Yes. Degree is defined exactly when order is. — Order always exists, just read off the highest derivative present. Degree needs one extra condition, being a polynomial in its derivatives, which is not automatic.
✓ No. Degree needs the equation to be a polynomial in its derivatives. — (B) Order only asks which derivative is highest, always answerable. Degree asks for that derivative’s power once the equation is a polynomial in its derivatives, which does not always hold.
No. Degree is undefined whenever more than one derivative term is present. — Multiple derivative terms are completely ordinary. What breaks degree is a derivative sitting inside a non-polynomial function like $\sin$ or $\ln$, not how many terms there are.
Yes. But only after the equation is solved for $y$. — Order and degree are read straight off the equation as given, with no solving required. Solving comes later and changes neither count.
A general solution of an order-2 differential equation has 2 arbitrary constants. Only ONE condition (one point the curve passes through) is given, not two. What do you get?
The complete particular solution: one condition is always enough.
Still the general solution: one condition changes nothing yet.
No solution at all: the constant count and the condition count do not match.
A smaller family than before: one constant fixed, one still free.
Check your answer
The complete particular solution: one condition is always enough. — One condition fixes one constant. With two constants in the general solution, one condition alone still leaves one free, not yet a single curve.
Still the general solution: one condition changes nothing yet. — Each condition fixes a constant as soon as it is applied. One condition narrows the family right away, without waiting for the rest.
No solution at all: the constant count and the condition count do not match. — Nothing here is a contradiction. A shortfall in conditions just leaves the family narrowed part of the way, not all the way to one curve.
✓ A smaller family than before: one constant fixed, one still free. — (D) One condition fixes one of the two constants. What remains is a smaller, one-constant family, narrower than the general solution, short of a single particular solution.
(Beyond the board syllabus.) A family of curves $y = C x$ has one arbitrary constant $C$. Differentiating once gives $d y/d x = C$. Is $d y/d x = C$ already the differential equation of the family?
Yes. A single differentiation always eliminates the constant automatically.
Yes. The order (1) now matches the number of constants (1).
No. $C$ is still present. It must be eliminated using the original equation.
No. A second differentiation is required, no matter how many constants there are.
Check your answer
Yes. A single differentiation always eliminates the constant automatically. — Differentiating changes the equation but does not remove $C$ by itself. $d y/d x = C$ still contains it, elimination is a separate algebraic step.
Yes. The order (1) now matches the number of constants (1). — The matching count (one differentiation, one constant) shows enough differentiating has been DONE. It does not mean $C$ has actually left the equation.
✓ No. $C$ is still present. It must be eliminated using the original equation. — (C) $d y/d x = C$ still contains $C$. The original equation gives $C = y/x$, so substituting yields $x ((d y)/(d x)) = y$, only this final, constant-free equation counts as the family’s differential equation.
No. A second differentiation is required, no matter how many constants there are. — The number of differentiations needed equals the number of constants. Here that is one, so a second differentiation is not required.
Exercise 9.2 practises checking a solution. Put the function into the equation, and see whether both sides agree. Then say whether it is general or particular, by counting its arbitrary constants.
Exercise 9.2: General and particular solutions
practice Verify that $y = e^{2 x}$ is a solution of $d y/d x - 2 y = 0$. Is it a general or a particular solution?
practice Verify that $y = x^3 - 3 x + C$ is a solution of $d y/d x = 3 x^2 - 3$. Is it a general or a particular solution?
practice Verify that $y = e^{-x} + x$ is a solution of $d y/d x + y = x + 1$. Is it a general or a particular solution?
practice Verify that $y = \ln(x) + C$ is a solution of $x d y/d x = 1$. Is it a general or a particular solution?
practice Verify that $x^2 + y^2 = 25$ solves $x + y d y/d x = 0$, for $-5 < x < 5$. Is it a general or a particular solution?
practice Verify that $y = x \tan^{-1}(x)$ solves $d y/d x = \tan^{-1}(x) + x/(1 + x^2)$. Is it a general or a particular solution?
practice Verify that $y = A \cos(2 x) + B \sin(2 x)$ solves $(d^2 y)/(d x^2) + 4 y = 0$, for arbitrary constants $A$ and $B$. Is it a general or a particular solution?
practice Verify that $y^2 = 4 a x$, for an arbitrary constant $a$, solves $2 x d y/d x = y$. Is it a general or a particular solution?
practice The number of arbitrary constants in the general solution of a differential equation of order $5$ is
$4$
$5$
$6$
$0$
practice The number of arbitrary constants in a particular solution of a differential equation of order $3$ is
$3$
$2$
$1$
$0$
practice Verify that $y = A e^{3 x} + B e^{-3 x}$ solves $(d^2 y)/(d x^2) - 9 y = 0$, for arbitrary constants $A$ and $B$. Is it a general or a particular solution?
practice Verify that $y = e^x (A \cos x + B \sin x)$ solves $(d^2 y)/(d x^2) - 2 d y/d x + 2 y = 0$, for arbitrary constants $A$ and $B$. Is it a general or a particular solution?
Answers
Verified: $d y/d x = 2 e^{2 x} = 2 y$, so $d y/d x - 2 y = 0$. Particular: the function carries no arbitrary constant, while the equation is order $1$.
Verified: $d y/d x = 3 x^2 - 3$ directly. General: it carries one arbitrary constant, $C$, matching the equation’s order $1$.
Verified: $d y/d x = -e^{-x} + 1$, so $d y/d x + y = -e^{-x} + 1 + e^{-x} + x = x + 1$. Particular: no arbitrary constant is present.
Verified: $d y/d x = 1/x$, so $x d y/d x = 1$. General: one arbitrary constant, matching order $1$.
Verified: differentiating implicitly, $2 x + 2 y d y/d x = 0$, so $x + y d y/d x = 0$. Particular: the constant $25$ is fixed, not arbitrary.
Verified: by the product rule, $d y/d x = \tan^{-1}(x) + x \cdot 1/(1+x^2)$, matching the right side exactly. Particular: no arbitrary constant is present.
Verified: $(d^2 y)/(d x^2) = -4 A \cos(2 x) - 4 B \sin(2 x) = -4 y$, so $(d^2 y)/(d x^2) + 4 y = 0$. General: two arbitrary constants, matching the equation’s order $2$.
Verified: differentiating implicitly, $2 y d y/d x = 4 a$, so $d y/d x = 2 a/y$; then $2 x d y/d x = 4 a x/y = y^2/y = y$, using $y^2 = 4 a x$. General: one arbitrary constant, matching order $1$.
(B) $5$.
(D) $0$.
Verified: $(d^2 y)/(d x^2) = 9 A e^{3 x} + 9 B e^{-3 x} = 9 y$, so $(d^2 y)/(d x^2) - 9 y = 0$. General: two arbitrary constants, matching order $2$.
Verified: differentiating twice and substituting confirms $(d^2 y)/(d x^2) - 2 d y/d x + 2 y = 0$ identically. General: two arbitrary constants, matching order $2$.
Exercise 9.3 practises separating the variables. Some items give a starting value, so finish with a particular solution. The word problems on money, population, bacteria and decay all use $d y/d t = k y$.
Exercise 9.3: Variables separable
practice Find the general solution of $d y/d x = x^2/y^2$.
practice Find the general solution of $d y/d x = (1+x)/(1+y)$.
practice Find the general solution of $d y/d x = e^{x-y}$.
practice Find the general solution of $2 y d y/d x = \sec^2(x)$.
practice Solve $d y/d x = 2 x y$, given that $y = 1$ when $x = 0$.
practice Solve $d y/d x = y/x$ ($x > 0$), given that $y = 2$ when $x = 1$.
practice A curve passes through the point $(0, 1)$. At every point $(x, y)$ on the curve, the slope of the tangent equals $x/y$. Find the equation of the curve.
practice A curve passes through the point $(0, 1)$. At every point $(x, y)$ on the curve, the slope of the tangent equals $-x/y$. Find the equation of the curve.
practice Find the general solution of $d y/d x = x y^2$.
practice Find the general solution of $x d y/d x = 1 + y^2$.
practice Solve $d y/d x = y \tan(x)$, given that $y = 3$ when $x = 0$.
practice Solve $d y/d x = (2 x)/(y+1)$, given that $y = 1$ when $x = 0$.
practice A curve passes through the point $(1, 2)$ ($x > 0$). At every point $(x, y)$ on the curve, the slope of the tangent equals $2 y/x$. Find the equation of the curve.
practice A curve passes through the point $(0, 1)$. At every point $(x, y)$ on the curve, the slope of the tangent equals $-(y+1)/(x+1)$. Find the equation of the curve.
practice Find the general solution of $d y/d x = (1+y^2)/(1+x^2)$.
practice A population grows at a rate proportional to its own size. If it doubles every $10$ years, express the population $P$ at time $t$ years in terms of the initial population $P_0$.
practice ₹$1000$ is deposited in an account where the amount grows continuously at $5\%$ per year, proportional to the amount present. Find the amount $A$ after $t$ years.
practice Find the general solution of $d y/d x + \sqrt{(1-y^2)/(1-x^2)} = 0$.
practice Solve $\cos(y) d y/d x = \sin(x)$, given that $y = 0$ when $x = 0$.
practice A radioactive substance decays at a rate proportional to the amount present. If $30\%$ of the initial amount decays in $5$ years, what percentage of the initial amount remains after $10$ years?
practice The number of bacteria in a culture grows at a rate proportional to the number present. If the count triples in $4$ hours, find the count after $12$ hours, as a multiple of the initial count.
practice The general solution of the differential equation $d y/d x = e^{x+y}$ is
$e^x + e^{-y} = C$
$e^x - e^y = C$
$e^{-x} + e^y = C$
$e^x + e^y = C$
practice Find the general solution of $(x y^2 + x) \mathrm{d} x + (y x^2 + y) \mathrm{d} y = 0$.
practice Solve $e^x \tan(y) \mathrm{d} x + (1-e^x) \sec^2(y) \mathrm{d} y = 0$, given that $y = \pi/4$ when $x = \ln(2)$.
Answers
$y^3 - x^3 = K$.
$y^2 + 2 y - x^2 - 2 x = K$.
$e^y = e^x + C$.
$y^2 = \tan(x) + C$.
$A = 1$; the particular solution is $y = e^{x^2}$.
$K = 2$; the particular solution is $y = 2 x$.
$y^2 - x^2 = 1$.
$x^2 + y^2 = 1$; through $(0, 1)$ the curve is the upper semicircle $y = \sqrt{1 - x^2}$.
$-1/y = x^2/2 + C$; $y = 0$ is also a solution, which this family does not contain.
$\tan^{-1}(y) = \ln \left|x\right| + C$.
$A = 3$; the particular solution is $y = 3 \sec(x)$.
$C = 2$; the particular solution is $(y+1)^2 = 2 x^2 + 4$.
$K = 2$; the curve is $y = 2 x^2$.
$K = 2$; the curve is $(x+1)(y+1) = 2$.
$\tan^{-1}(y) = \tan^{-1}(x) + C$.
$P = P_0 \cdot 2^{t/10}$.
$A = 1000 e^{0.05 t}$.
$\sin^{-1}(y) + \sin^{-1}(x) = C$, for $\left|x\right| < 1$ and $\left|y\right| < 1$; $y = 1$ and $y = -1$ are also solutions, which this family does not contain.
$C = 1$; the particular solution is $\sin(y) = 1 - \cos(x)$, so $y = \sin^{-1}(1 - \cos(x))$ near $x = 0$.
$49\%$ of the initial amount remains after $10$ years.
$27$ times the initial count.
(A) $e^x + e^{-y} = C$.
$(x^2+1)(y^2+1) = K$.
$K = -1$; the particular solution is $\tan(y) = e^x - 1$.
Exercise 9.4 practises homogeneous equations. Test each one with $\lambda$ first. After you put $y = v x$, finish the algebra before you separate.
Exercise 9.4: Homogeneous equations
practice Show that $d y/d x = (x-y)/x$ is homogeneous, and solve it.
practice Show that $d y/d x = (2x+y)/x$ is homogeneous, and solve it.
practice Show that $d y/d x = (x^2+y^2)/(x y)$ is homogeneous, and solve it.
practice Show that $d y/d x = (x+y)/(2x)$ is homogeneous, and solve it.
practice Show that $d y/d x = (x+3y)/(2x)$ is homogeneous, and solve it.
practice Show that $d y/d x = y/x + \tan(y/x)$ is homogeneous, and solve it.
practice Show that $d y/d x = (x^2+y^2)/(2 x y)$ is homogeneous, and solve it.
practice Show that $d y/d x = (x+y)/(x-y)$ is homogeneous, and solve it.
practice Solve $d y/d x = (x-y)/x$, given that $y = 1$ when $x = 1$.
practice Solve $d y/d x = (2x+y)/x$, given that $y = 0$ when $x = 1$.
practice Solve $d y/d x = (x+y)/(2x)$, given that $y = 3$ when $x = 1$.
practice Solve $d y/d x = (x^2+y^2)/(x y)$, given that $y = 2$ when $x = 1$.
practice Solve $d y/d x = (x^2+y^2)/(2 x y)$, given that $y = 1$ when $x = 2$.
practice Solve $d y/d x = (x+3y)/(2x)$, given that $y = 1$ when $x = 1$.
practice A curve passes through the point $(4, 2)$ ($x > 0$). At every point $(x, y)$ on it, the slope of the tangent equals $(x+y)/(2x)$. Find the equation of the curve.
practice For a homogeneous equation written as $d x/d y = h(x/y)$, which substitution gives an equation in $v$ and $y$ with its variables separable?
$x = v y^2$
$v = x y$
$x = v y$
$x = v$
practice Substituting $y = v x$ in the homogeneous equation $d y/d x = y/x + \tan(y/x)$ turns it into
$x d v/d x = \tan(v)$
$x d v/d x = v \tan(v)$
$d v/d x = \tan(v)$
$v d v/d x = \tan(x)$
practice Solve $(x^2 - y^2) \mathrm{d} x + 2 x y \mathrm{d} y = 0$.
practice Solve $(1 + 2 e^{x/y}) \mathrm{d} x + 2 e^{x/y} (1 - x/y) \mathrm{d} y = 0$.
practice Which of these is homogeneous? (a) $(x^2 + x y) \mathrm{d} y = (x^2 + y^2) \mathrm{d} x$ (b) $d y/d x = (x + y + 1)/(x - y)$
Answers
$y = x/2 + C/x$.
$y = 2 x \ln \left|x\right| + C x$.
$y^2 = x^2 (2 \ln \left|x\right| + C)$.
$y = x + C \sqrt{\left|x\right|}$.
$y = K x \sqrt{\left|x\right|} - x$.
$\sin(y/x) = K x$.
$x^2 - y^2 = K x$.
$\tan^{-1}(y/x) - (1/2) \ln(x^2+y^2) = C$.
$C_1 = 1/2$; the particular solution is $y = x/2 + 1/(2x)$.
$C = 0$; the particular solution is $y = 2 x \ln(x)$.
$C_1 = 2$; the particular solution is $y = x + 2 \sqrt{x}$.
$C_1 = 4$; the particular solution is $y^2 = x^2 (2 \ln(x) + 4)$.
$K = 3/2$; the particular solution is $x^2 - y^2 = (3/2) x$.
$K = 2$; the particular solution is $y = 2 x \sqrt{x} - x$.
$C_1 = -1$; the curve is $y = x - \sqrt{x}$.
(C) $x = v y$.
(A) $x d v/d x = \tan(v)$.
$x^2 + y^2 = K x$.
$x + 2 y e^{x/y} = K$.
(a) is homogeneous; (b) is not, because the $1$ does not scale.
Exercise 9.5 practises linear equations and the integrating factor. Write each one in the form $d y/d x + P y = Q$ first. Some items have $x$ as the dependent variable, so they use $d x/d y + P_1 x = Q_1$.
Exercise 9.5: Linear equations
practice Find the general solution of $d y/d x + y = e^x$.
practice Find the general solution of $d y/d x + 2 y = x$.
practice Find the general solution of $d y/d x - y = x$.
practice Find the general solution of $x d y/d x + 2 y = x^2$.
practice Find the general solution of $d y/d x + y \cot(x) = \cos(x)$.
practice Find the general solution of $\cos^2(x) d y/d x + y = \tan(x)$.
practice Find the general solution of $d y/d x + y \tan(x) = \sin(x)$.
practice Find the general solution of $d y/d x + (2x)/(1+x^2) y = 1/(1+x^2)^2$.
practice Find the general solution of $d y/d x + y = \cos(x)$.
practice Find the general solution of $d y/d x + 4 y = e^{-x}$.
practice Find the general solution of $d x/d y + x = y$, with $x$ as the dependent variable.
practice Find the general solution of $d x/d y + (2/y) x = y^2$ ($y > 0$), with $x$ as the dependent variable.
practice Solve $d y/d x + y = e^x$, given that $y = 1$ when $x = 0$.
practice Solve $d y/d x + 2 y = x$, given that $y = 0$ when $x = 0$.
practice Solve $d y/d x - y = x$, given that $y = 0$ when $x = 0$.
practice Solve $x d y/d x + 2 y = x^2$, given that $y = 1$ when $x = 1$.
practice Solve $d x/d y + x = y$, given that $x = 0$ when $y = 0$, with $x$ as the dependent variable.
practice A curve passes through the point $(0, 1)$. At every point $(x, y)$ on the curve, the slope of the tangent plus $y$ equals $x$. Find the equation of the curve.
practice A curve passes through the point $(1, 0)$ ($x > 0$). At every point $(x, y)$ on the curve, the slope of the tangent minus $y/x$ equals $x$. Find the equation of the curve.
practice An integrating factor of $x d y/d x - y = 2 x^2$ is
$e^{-x}$
$1/x$
$x$
$e^x$
practice An integrating factor of the differential equation $d y/d x + y \tan(x) = \sec(x)$ is
$\cos(x)$
$\sin(x)$
$\sec(x)$
$\tan(x)$
practice Solve $y \mathrm{d} x - (x + 2 y^2) \mathrm{d} y = 0$.
Answers
$y = e^x/2 + C e^{-x}$.
$y = x/2 - 1/4 + C e^{-2x}$.
$y = -(x+1) + C e^x$.
$y = x^2/4 + C/x^2$.
$y = \sin(x)/2 + C/\sin(x)$.
$y = \tan(x) - 1 + C e^{-\tan(x)}$.
$y = \cos(x) (C - \ln|\cos(x)|)$.
$y = (\tan^{-1}(x) + C)/(1+x^2)$.
$y = (\cos(x) + \sin(x))/2 + C e^{-x}$.
$y = e^{-x}/3 + C e^{-4x}$.
$x = y - 1 + C e^{-y}$.
$x = y^3/5 + C/y^2$.
$C = 1/2$; the particular solution is $y = e^x/2 + e^{-x}/2$.
$C = 1/4$; the particular solution is $y = x/2 - 1/4 + e^{-2x}/4$.
$C = 1$; the particular solution is $y = e^x - x - 1$.
$C = 3/4$; the particular solution is $y = x^2/4 + 3/(4x^2)$.
$C = 1$; the particular solution is $x = y - 1 + e^{-y}$.
$C = 2$; the curve is $y = x - 1 + 2 e^{-x}$.
$C = -1$; the curve is $y = x^2 - x$.
(B) $1/x$.
(C) $\sec(x)$.
$x = 2 y^2 + C y$; $y = 0$ is also a solution, which this family does not contain.
The Miscellaneous Exercise mixes every method, and it does not say which one to use. So first decide whether each equation separates, is homogeneous or is linear. It ends with growth, decay and a mixing problem.
Miscellaneous Exercise: Chapter 9
practice Find the order and degree, if defined, of the differential equation $(d^2 y)/(d x^2) + 3 x ((d y)/(d x))^2 - 4 y = e^x$.
practice Find the order and degree, if defined, of the differential equation $(d^3 y)/(d x^3) + \cos((d^3 y)/(d x^3)) = x^2$.
practice Verify that $y = 3 \cos(2 x) - 4 \sin(2 x)$ solves $(d^2 y)/(d x^2) + 4 y = 0$. Is it a general or a particular solution?
practice Verify that $x^3 - 3 x y^2 = c$, for an arbitrary constant $c$, solves $d y/d x = (x^2-y^2)/(2 x y)$.
practice Solve $d y/d x = (1+y^2)/(1+x^2)$, given that $y = 1$ when $x = 0$, and write the answer without an inverse trig function.
practice Show that $d y/d x = (x+y)/(2x)$ is homogeneous, and find its general solution.
practice Find the general solution of $d y/d x + y = \sin(x)$.
practice The number of bacteria in a culture grows at a rate proportional to the number present. If the count doubles in $3$ hours, find the count after $9$ hours, as a multiple of the initial count.
practice A bank pays interest continuously at $8\%$ per year, proportional to the amount present. If ₹$5000$ is deposited, find the amount after $10$ years, in terms of $e$.
practice Verify that $y = A e^{2 x} + B e^{-2 x} + x$ solves $(d^2 y)/(d x^2) - 4 y = -4 x$, for arbitrary constants $A$ and $B$. Is it a general or a particular solution?
practice Solve $d y/d x - y \tan(x) = e^x \sec(x)$, given that $y = 1$ when $x = 0$.
practice Solve $(x^2-y^2) \mathrm{d} x + 2 x y \mathrm{d} y = 0$, given that $y = 1$ when $x = 1$.
practice Which method solves $d y/d x = (x^2+y^2)/(x y)$?
variables separable
homogeneous, substitute $y = v x$
linear, use an integrating factor
none of these
practice An integrating factor of $d y/d x + (2/x) y = x^3$ is
$x$
$x^2$
$1/x^2$
$x^3$
practice Solve $d y/d x = y + 2 x$.
practice Solve $d y/d x = (x+y)/(x-y)$, given that $y = 0$ when $x = 1$.
practice A radioactive substance decays at a rate proportional to the amount present. If a sample takes 25 years to reduce to half its original mass, find the fraction of the original mass remaining after 75 years.
practice A mixture-style word problem: a tank holds 100 litres of brine with 5 kg of dissolved salt. Fresh brine carrying 0.2 kg of salt per litre flows in at 4 litres per minute, and the stirred mixture drains out at the same rate. If $y$ is the salt present (in kg) after $t$ minutes, find $y$ as a function of $t$.
Answers
Order $2$, degree $1$.
Order $3$, degree not defined.
Verified: $(d^2 y)/(d x^2) = -12 \cos(2 x) + 16 \sin(2 x) = -4 y$, so $(d^2 y)/(d x^2) + 4 y = 0$. Particular: no arbitrary constant is present.
Verified: differentiating implicitly, $3 x^2 - 3 y^2 - 6 x y d y/d x = 0$, which rearranges to $d y/d x = (x^2-y^2)/(2 x y)$.
$y = (x+1)/(1-x)$.
$y = x + C \sqrt{\left|x\right|}$.
$y = (\sin(x) - \cos(x))/2 + C e^{-x}$.
$8$ times the initial count.
$A = 5000 e^{0.8}$.
Verified: $(d^2 y)/(d x^2) - 4 y = 4 A e^{2 x} + 4 B e^{-2 x} - 4 A e^{2 x} - 4 B e^{-2 x} - 4 x = -4 x$. General: two arbitrary constants, matching order $2$.
$C = 0$; the particular solution is $y = e^x \sec(x)$.
$K = 2$; the particular solution is $x^2 + y^2 = 2 x$.
(B) homogeneous, substitute $y = v x$.
(B) $x^2$.
$y = -2 (x+1) + C e^x$.
$C = 0$; the particular solution is $\tan^{-1}(y/x) = (1/2) \ln(x^2+y^2)$.
$1/8$ of the original mass.
$y = 20 - 15 e^{-0.04 t}$.
Board paper practice
practice The order and degree of the differential equation $((d^2 y)/(d x^2))^3 + ((d y)/(d x))^4 + \sin(x) = 0$ are ______. (1 mark)
Order $2$, degree $4$
Order $2$, degree $3$
Order $1$, degree $4$
Order $2$, degree not defined
practice Assertion (A). The degree of the differential equation $(d^2 y)/(d x^2) + \sin((d y)/(d x)) = 0$ is not defined. Reason (R). The equation is not a polynomial in its derivatives. (1 mark)
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
practice Solve $d y/d x = x e^{-y}$, given that $y = 0$ when $x = 0$. (2 marks)
practice Show that $x^2 d y/d x = x^2 + x y + y^2$ is homogeneous, and find its general solution. (3 marks)
practice Solve $(1 + y^2) \mathrm{d} x = (\tan^{-1}(y) - x) \mathrm{d} y$, given that $x = 0$ when $y = 0$. (5 marks)
practice A town’s population $P$ grows at a rate proportional to its size: $d P/d t = k P$, with $t$ in years. In 2010 the town had 50,000 people, and in 2020 it had 60,000. Take $t = 0$ in 2010. (i) Write $P$ in terms of $t$ and $k$. (1 mark) (ii) Find $k$. (1 mark) (iii) Find the population in 2030. (2 marks)
Answers
(B) order $2$, degree $3$.
Both A and R are true and R is the correct explanation of A (order $2$, degree not defined).
$y = \ln((x^2 + 2)/2)$.
$\tan^{-1}(y/x) = \ln \left|x\right| + C$.
$x = \tan^{-1}(y) - 1 + e^{-\tan^{-1}(y)}$.
(i) $P = 50000 e^{k t}$. (ii) $k = \ln(1.2)/10 \approx 0.0182$ per year. (iii) 72,000 people.
