Chapter 9 · Differential Equations · Go for it

Differential equations, the quick way

A differential equation is a rule about a slope. Read off its order and degree, then pick one of three methods to solve it. Nine kinds of question cover the chapter.

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The chapter in one minute

What a differential equation is, and what the exam asks.

A differential equation is an equation with a derivative in it, like $\frac{dy}{dx} = x + y$ ⁠. It is a rule about the slope of a curve.

To solve it, you find $y$ itself. The general solution has a constant $C$ in it. A particular solution has a number in place of $C$ ⁠, found from a given point.

The exam asks for the order and degree, asks you to check a solution, and asks you to solve an equation by one of three methods: separate the variables, put $y = vx$ ⁠, or use an integrating factor.

Most questions are one of nine kinds, and each kind has a fixed recipe. You will learn all nine below.

9question types

12formula cards

3ways to solve: separate, put y = vx, integrating factor

1check that catches most mistakes: put your answer back into the equation

Formulas

The formulas you must know

Learn these cards. Every question in the chapter is built from them.

Order and degree

\left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} = 0

Order 2: the highest derivative is the second one. Degree 3: the highest power of that derivative.

Degree not defined

\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0

A degree needs a polynomial in the derivatives. A derivative inside sin, cos, tan, log or e is not one, so the degree is not defined. The order is still 1.

General solution

y = a\cos x + b\sin x

It solves $y'' + y = 0$ ⁠, which has order 2, so it has 2 constants, $a$ and $b$ ⁠. A particular solution has none.

Variables separable

\begin{aligned} &\frac{dy}{dx} = g(x)\,h(y) \\ &\Rightarrow \int\frac{dy}{h(y)} = \int g(x)\,dx \end{aligned}

All $y$ with $dy$ ⁠, all $x$ with $dx$ ⁠. Then integrate and add $+\,C$ ⁠.

Homogeneous test

F(\lambda x, \lambda y) = F(x, y)

If this holds, $\frac{dy}{dx} = F(x, y)$ is homogeneous. Quick look: every term on the top and bottom of the fraction has the same degree, or $y$ and $x$ appear only as $\frac{y}{x}$ ⁠.

Put

y = vx

\begin{aligned} y &= vx \\ \frac{dy}{dx} &= v + x\frac{dv}{dx} \end{aligned}

Then separate $v$ and $x$ ⁠. At the end, put $v = \frac{y}{x}$ back.

Linear equation

\frac{dy}{dx} + Py = Q

$P$ and $Q$ are numbers or functions of $x$ only. Divide first, so that nothing multiplies $\frac{dy}{dx}$ ⁠.

Integrating factor

\text{I.F.} = e^{\int P\,dx}

Integrate $P$ first, then raise $e$ to it.

The linear answer

\begin{aligned} &y\cdot\text{I.F.} \\ &= \int Q\cdot\text{I.F.}\,dx + C \end{aligned}

Multiply $Q$ by the I.F., integrate, and add $C$ ⁠.

x

as the unknown

\frac{dx}{dy} + P_1x = Q_1

$P_1$ and $Q_1$ are numbers or functions of $y$ only. Then $\text{I.F.} = e^{\int P_1\,dy}$ and $x\cdot\text{I.F.}$ $= \int Q_1\cdot\text{I.F.}\,dy + C$ ⁠.

e and log cancel

e^{\log f(x)} = f(x)

For $f(x) > 0$ ⁠. So $e^{\log x} = x$ and $e^{-\log x} = \frac{1}{x}$ ⁠.

Growth at a rate

\begin{aligned} &\frac{dP}{dt} = kP \\ &\Rightarrow P = P_0e^{kt} \end{aligned}

$P_0$ is the amount at the start. For $r$ % a year, $k = \frac{r}{100}$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Order and degree: the highest derivative, and its power

Spot it: "find the order and degree (if defined)". Often an MCQ.

Find the highest derivative. If it is $y'''$ ⁠, the order is 3.
Check that the equation is a polynomial in the derivatives. A derivative inside sin, cos, tan, log or e is not, so then the degree is not defined.
If it is a polynomial, the degree is the highest power of the highest derivative. Powers of other terms do not count.

Exercise 9.1 Q6. Find the order and degree of

\begin{aligned} &(y''')^2 + (y'')^3 \\ &\quad + (y')^4 + y^5 = 0 \end{aligned}

The highest derivative is $y'''$ ⁠, so the order is 3.

It is a polynomial in $y'''$ ⁠, $y''$ and $y'$ ⁠, so the degree is defined. It is the highest power of $y'''$ ⁠, which is 2. The 5 on $y$ does not count: $y$ is not a derivative.

∴ Order 3; degree 2

See a full exam answer · Book Example 1 · Go deeper: see it visually · Basics: y′ and y″

In past papers (8): click to see

Board 2026 Q12 · Board 2024 Q14 · Board 2023 Q7 · Board 2022 T2 Q1 · Sample 2026-27 Q11 · Sample 2023-24 Q6 · Sample 2022-23 Q6 · Sample 2021-22 T2 Q2

2. Is it a solution? Differentiate, then substitute

Spot it: "verify that the given function is a solution of the differential equation".

Differentiate the function as many times as the equation needs.
Put $y$ ⁠, $y'$ (and $y''$ ⁠) into each side, and simplify.
Show that L.H.S. equals R.H.S., and say so in words.

Exercise 9.2 Q4. Verify that $y = \sqrt{1 + x^2}$ is a solution of $y' = \frac{xy}{1 + x^2}$ ⁠.

\begin{aligned} \text{L.H.S.} = y' &= \frac{2x}{2\sqrt{1 + x^2}} \\ &= \frac{x}{\sqrt{1 + x^2}} \end{aligned}

\begin{aligned} \text{R.H.S.} &= \frac{x\sqrt{1 + x^2}}{1 + x^2} \\ &= \frac{x}{\sqrt{1 + x^2}} \end{aligned}

∴ $\text{L.H.S.} = \text{R.H.S.}$ ⁠, so the function is a solution

See a full exam answer · Book Examples 2 and 3 · Go deeper: see it visually

3. Variables separable: $y$ on one side, $x$ on the other

Spot it: $\frac{dy}{dx}$ equals a part in $x$ times a part in $y$ ⁠, like $(1 + x^2)(1 + y^2)$ ⁠.

Move every $y$ to the side with $dy$ ⁠, and every $x$ to the side with $dx$ ⁠.
Put an integral sign on both sides and integrate.
Add one $+\,C$ ⁠, on the $x$ side.

Exercise 9.3 Q6. Solve $\frac{dy}{dx} = (1 + x^2)(1 + y^2)$ ⁠.

Separate the variables:

\frac{dy}{1 + y^2} = (1 + x^2)\,dx

Integrate both sides:

\begin{aligned} &\int\frac{dy}{1 + y^2} \\ &= \int(1 + x^2)\,dx \end{aligned}

∴ $\tan^{-1}y = x + \frac{x^3}{3} + C$

See a full exam answer · Book Examples 4 to 9 · Go deeper: see it visually · Basics: dy and dx

In past papers (7): click to see

Board 2026 Q29(b) · Board 2023 Q15 · Sample 2025-26 Q10 · Sample 2024-25 Q14 · Sample 2023-24 Q15 · Sample 2022-23 Q15 · Sample 2020-21 Q25

4. A particular solution: solve, then find $C$

Spot it: a condition such as " $y = 1$ when $x = 0$ ", or "the curve passes through a point".

Find the general solution, with $+\,C$ ⁠.
Put in the given $x$ and $y$ ⁠, on a line of their own, and work out $C$ ⁠.
Put that value of $C$ back. The answer has no $C$ ⁠.

Exercise 9.3 Q14. Solve $\frac{dy}{dx} = y\tan x$ ⁠, given $y = 1$ when $x = 0$ ⁠.

Separate and integrate:

\frac{dy}{y} = \tan x\,dx

\log|y| = \log|\sec x| + C

Put $x = 0$ and $y = 1$ ⁠: $\log 1 = \log 1 + C$ ⁠, so $C = 0$ ⁠.

So $|y| = |\sec x|$ ⁠. At the given point both are positive.

∴ $y = \sec x$

See a full exam answer · Book Examples 6 and 7 · Go deeper: see it visually · Basics: one constant per integration

In past papers (10): click to see

Board 2026 Q29(b) · Board 2024 Q29 · Board 2023 Q28(a) · Board 2022 T2 Q10 · Sample 2026-27 Q29(a) · Sample 2026-27 Q29(b) · Sample 2025-26 Q34(b) · Sample 2021-22 T2 Q8(b) · Sample 2020-21 Q9(a) · Sample 2020-21 Q25

5. Growth at a rate: money and bacteria

Spot it: "increases continuously at 5% per year", or "the rate of growth is proportional to the number present".

Write the rate as an equation: $\frac{dP}{dt} = kP$ ⁠.
Separate and integrate: $P = Ae^{kt}$ ⁠.
Use the start value to find $A$ ⁠, then put in the time asked for.

Exercise 9.3 Q21. Money grows continuously at 5% per year. What does Rs 1000 become after 10 years? Use $e^{0.5} = 1.648$ ⁠.

Let $P$ be the amount after $t$ years. It grows at 5% of itself each year:

\frac{dP}{dt} = \frac{5P}{100} = \frac{P}{20}

Separate and integrate:

\int\frac{dP}{P} = \int\frac{dt}{20}

\log P = \frac{t}{20} + C

So $P = Ae^{t/20}$ ⁠, where $A = e^C$ ⁠. At the start, $t = 0$ and $P = 1000$ ⁠, so $A = 1000$ ⁠.

\begin{aligned} P &= 1000e^{10/20} \\ &= 1000\times 1.648 = 1648 \end{aligned}

∴ After 10 years, $P = 1648$ ⁠: the amount is Rs 1648

See a full exam answer · Book Example 9 · Go deeper: see it visually · Basics: growth and e

6. Homogeneous equations: put $y = vx$

Spot it: every term has the same degree, like $x^2$ ⁠, $xy$ and $y^2$ ⁠, or you see $\frac{y}{x}$ inside sin, cos or log. You cannot separate $x$ and $y$ ⁠.

Write $\frac{dy}{dx} = F(x, y)$ and show that $F(\lambda x, \lambda y) = F(x, y)$ ⁠.
Put $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠.
Separate $v$ and $x$ ⁠, and integrate.
Put $v = \frac{y}{x}$ back and tidy up.

Exercise 9.4 Q4. Solve $(x^2 - y^2)\,dx + 2xy\,dy = 0$ ⁠.

\frac{dy}{dx} = \frac{y^2 - x^2}{2xy} = F(x, y)

\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda^2(y^2 - x^2)}{\lambda^2\cdot 2xy} \\ &= F(x, y) \end{aligned}

So it is homogeneous. Put $y = vx$ ⁠, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠.

\begin{aligned} v + x\frac{dv}{dx} &= \frac{v^2 - 1}{2v} \\ x\frac{dv}{dx} &= -\frac{1 + v^2}{2v} \end{aligned}

\frac{2v}{1 + v^2}\,dv = -\frac{dx}{x}

\begin{aligned} &\log(1 + v^2) \\ &= -\log|x| + \log C \end{aligned}

So $(1 + v^2)\,x = C$ ⁠. Put $v = \frac{y}{x}$ back:

\left(1 + \frac{y^2}{x^2}\right)x = C

∴ $x^2 + y^2 = Cx$

See a full exam answer · Book Examples 10 to 13 · Go deeper: see it visually · Basics: same degree

In past papers (12): click to see

Board 2026 Q29(a) · Board 2025 Q4 · Board 2024 Q29 · Board 2023 Q28(b) · Board 2022 T2 Q10 · Sample 2026-27 Q29(b) · Sample 2025-26 Q34(b) · Sample 2024-25 Q5 · Sample 2023-24 Q29(a) · Sample 2022-23 Q29(b) · Sample 2021-22 T2 Q8(a) · Sample 2020-21 Q9(b)

7. Linear equations: the integrating factor

Spot it: $\frac{dy}{dx} + Py = Q$ ⁠, with $P$ and $Q$ in $x$ only, and $y$ only to power 1.

Divide so that $\frac{dy}{dx}$ stands alone. Read off $P$ and $Q$ ⁠.
Find $\text{I.F.} = e^{\int P\,dx}$ and simplify it.
Write $y\cdot\text{I.F.}$ $= \int Q\cdot\text{I.F.}\,dx + C$ ⁠, and integrate.

Exercise 9.5 Q3. Solve $\frac{dy}{dx} + \frac{y}{x} = x^2$ ⁠.

It is linear, with $P = \frac{1}{x}$ and $Q = x^2$ ⁠.

\begin{aligned} \text{I.F.} &= e^{\int\frac{1}{x}dx} \\ &= e^{\log x} = x \end{aligned}

\begin{aligned} y\cdot x &= \int x^2\cdot x\,dx \\ &= \frac{x^4}{4} + C \end{aligned}

∴ $xy = \frac{x^4}{4} + C$

See a full exam answer · Book Examples 14 to 18 · Go deeper: see it visually · Basics: why the I.F. works

In past papers (7): click to see

Board 2024 Q8 · Board 2023 Q28(a) · Sample 2026-27 Q29(a) · Sample 2025-26 Q34(a) · Sample 2023-24 Q29(b) · Sample 2021-22 T2 Q8(b) · Sample 2020-21 Q35

8. Linear in $x$ ⁠: when $y$ will not behave, solve for $x$

Spot it: there is a $y^2$ or $\tan^{-1}y$ next to $dy$ ⁠, but $x$ appears only to power 1.

Divide to get $\frac{dx}{dy} + P_1x = Q_1$ ⁠, with $P_1$ and $Q_1$ in $y$ only.
Find $\text{I.F.} = e^{\int P_1\,dy}$ ⁠.
Write $x\cdot\text{I.F.}$ $= \int Q_1\cdot\text{I.F.}\,dy + C$ ⁠.

Exercise 9.5 Q11. Solve $y\,dx + (x - y^2)\,dy = 0$ ⁠.

Here $x$ appears only to power 1, so divide by $y\,dy$ and take $x$ as the unknown:

\frac{dx}{dy} + \frac{x}{y} = y

So $P_1 = \frac{1}{y}$ and $Q_1 = y$ ⁠.

\begin{aligned} \text{I.F.} &= e^{\int\frac{1}{y}dy} \\ &= e^{\log y} = y \end{aligned}

\begin{aligned} x\cdot y &= \int y\cdot y\,dy \\ &= \frac{y^3}{3} + C \end{aligned}

∴ $x = \frac{y^2}{3} + \frac{C}{y}$

See a full exam answer · Book Example 16 · Go deeper: see it visually

In past papers (3): click to see

Board 2026 Q11 · Board 2025 Q12 · Sample 2022-23 Q29(a)

9. A curve from words: turn the slope into an equation

Spot it: "the slope of the tangent at any point $(x, y)$ is …", and a point the curve passes through.

The slope of the tangent is $\frac{dy}{dx}$ ⁠. Write the words as an equation.
Decide the type (separable, homogeneous or linear) and solve it.
Use the point to find $C$ ⁠, as in type 4.

Exercise 9.5 Q16. Find the equation of a curve through the origin, given that the slope of the tangent at any point $(x, y)$ is the sum of the coordinates of the point.

\frac{dy}{dx} = x + y

\frac{dy}{dx} - y = x

It is linear, with $P = -1$ and $Q = x$ ⁠. So $\text{I.F.} = e^{\int -1\,dx} = e^{-x}$ ⁠.

\begin{aligned} ye^{-x} &= \int xe^{-x}\,dx \\ &= -xe^{-x} - e^{-x} + C \end{aligned}

The integral is by parts. Through the origin: put $x = 0$ and $y = 0$ ⁠: $0 = 0 - 1 + C$ ⁠, so $C = 1$ ⁠.

ye^{-x} = -xe^{-x} - e^{-x} + 1

Multiply by $e^x$ and rearrange.

∴ $x + y + 1 = e^x$

See a full exam answer · Book Example 18 · Go deeper: see it visually · Basics: a rule for slopes

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Check your answer: differentiate it and put it back into the equation. If both sides agree, you are right.
Pick the method in ten seconds. Can you split $x$ and $y$ ⁠? Separate. Same degree in every term? Put $y = vx$ ⁠. Does it fit $\frac{dy}{dx} + Py = Q$ ⁠? Integrating factor.
Order and degree MCQ: look first for a derivative inside sin, cos, tan, log or e. If there is one, the answer is "not defined".
But $\sin y$ is fine, because $y$ is not a derivative: $y'' + 2y' + \sin y = 0$ has order 2 and degree 1.
A general solution of an order 4 equation has 4 constants. A particular solution has none.
When the other terms are logs, write the constant as $\log C$ ⁠. Then the logs combine in one step.
Simplify the I.F. at once: $e^{\log x} = x$ ⁠, $e^{-\log x} = \frac{1}{x}$ ⁠, $e^{2\log x} = x^2$ ⁠.
Divide first, then read off $P$ ⁠: for $x\frac{dy}{dx} - y = 2x^2$ ⁠, divide by $x$ ⁠, so $P = -\frac{1}{x}$ and $\text{I.F.} = \frac{1}{x}$ ⁠.
Find $C$ straight after integrating, before you tidy up the answer.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Taking the degree from the biggest power anywhere → use the power of the highest derivative only.
Writing the solution of $\frac{dy}{dx} = y$ as $y = e^x + C$ → separate first: $\log|y| = x + C$ ⁠, so $y = Ae^x$ ⁠.
Using $\text{I.F.} = e^{P}$ → integrate first: $\text{I.F.} = e^{\int P\,dx}$ ⁠.
Reading off $P$ while $\frac{dy}{dx}$ still has a number or $x$ in front → divide through first.
Writing $y = \int Q\cdot\text{I.F.}\,dx + C$ → the left side is $y\cdot\text{I.F.}$ ⁠, not $y$ ⁠.
Leaving $v$ in the answer → put $v = \frac{y}{x}$ back.
Putting in the given point before adding $+\,C$ → add $C$ when you integrate, then use the point.

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

Marks go to value points: lines that change the form. Here that means the separated line, the $y = vx$ line with its $\frac{dy}{dx}$ ⁠, and the I.F. line. Write each one.
Say what you use, in a few words: "homogeneous, so put $y = vx$ ", or "linear, with $P = \ldots$ and $Q = \ldots$ ".
A general solution ends with one $+\,C$ ⁠. For a particular solution, show the line where the given values go in and $C$ comes out.
In a "verify" (show that) question, end with "L.H.S. = R.H.S., so it is a solution". CBSE's model answers give the conclusion line half a mark.
Give the final answer on its own line. A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 9.1 Q4 a derivative inside cos: degree not defined
Exercise 9.2 Q1 verify, with $y'$ and $y''$
Exercise 9.3 Q4 separate the variables
Exercise 9.3 Q15 a curve through $(0, 0)$ ⁠: find $C$
Exercise 9.5 Q18 MCQ: divide by $x$ first, then the I.F.
Exercise 9.5 Q2 a linear equation with $P = 3$
Exercise 9.5 Q12 linear in $x$ ⁠: take $x$ as the unknown
Exercise 9.4 Q3 homogeneous: put $y = vx$
Exercise 9.4 Q12 homogeneous, then find $C$
Example 18 a slope in words, then a linear equation
Exercise 9.3 Q22 bacteria: growth proportional to the number

You can do this

Nine kinds of question, one recipe each. Spot the type, solve, and put your answer back into the equation to check it. Work through the practice list and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

31 CBSE board and sample-paper questions on Differential Equations, each with a full-marks exam answer. Practise them →