MathsCBSE Class 12 · Chapter 9

Differential Equations Class 12: Notes, Methods and Solved Questions

Updated 14 October 2026 · Fact-checked

Differential Equations is Chapter 9 of CBSE Class 12 Maths. It covers the order and degree of a differential equation, general and particular solutions, and three methods for first-order, first-degree equations: separating the variables, the substitution y = vx for homogeneous equations, and the integrating factor for linear equations.

What Differential Equations is about

A differential equation is an equation that contains derivatives of an unknown function, such as $\frac{d y}{d x} = 2 x y$ or $y^{''} + y = 0$ . Many laws of science are written this way: the rate at which a population grows, a cup of tea cools or a medicine leaves the blood is proportional to what is present. Solving the equation means finding the function itself.

You first learn to describe an equation by its order (the highest derivative present) and degree (the power of that highest derivative, defined only when the equation is a polynomial in its derivatives). A general solution contains as many arbitrary constants as the order; a particular solution fixes them using given conditions such as y(0) = 1.

For first-order, first-degree equations the syllabus has three methods: variables separable (bring all y's to one side and all x's to the other, then integrate), homogeneous equations (put y = vx), and linear equations $\frac{d y}{d x} + P y = Q$ or $\frac{d x}{d y} + P_{1} x = Q_{1}$ (multiply by the integrating factor). Forming a differential equation from a given general solution is not in the 2026-27 syllabus.

Calculus carries 35 of the 80 marks. In our question bank of eight board papers (2023 to 2026) and four sample papers, this chapter has 38 questions and appears in all eight board papers; order and degree MCQs and 3-mark homogeneous or linear equations account for most of them.

Weightage: This chapter is part of Unit III, Calculus, which carries 35 of the 80 theory marks together with Continuity and Differentiability, Application of Derivatives, Integrals and Application of Integrals.Source: CBSE Mathematics syllabus 2026-27, Unit III Calculus

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.4 marks per paper. See the questions

Key concepts

Differential equations and their order

NCERT section 9.2.1

An equation involving the derivatives of a dependent variable y with respect to one independent variable x is an ordinary differential equation (the only kind in this chapter). Examples: $\frac{d y}{d x} = cos x$ , $y^{''} + 4 y = 0$ .

The order is the order of the highest derivative that appears. It is always a positive integer. $\frac{d y}{d x} + y = e^{x}$ has order 1; $\frac{d ^{3} y}{d x ^{3}} + (\frac{d y}{d x})^{5} = 0$ has order 3, even though the first derivative has the bigger power.

Worked example

Find the order of $\frac{d ^{3} y}{d x ^{3}} + 2 (\frac{d ^{2} y}{d x ^{2}})^{4} + y = sin x$ . [1 mark]

Try it first, then show the solutionSolution

Highest derivative present: $\frac{d ^{3} y}{d x ^{3}}$ .

Answer: Order 3

Degree of a differential equation

NCERT section 9.2.2

When the equation is a polynomial in its derivatives, its degree is the highest power (a positive integer) of the highest-order derivative, after clearing fractional powers.

Remove roots or fractional powers of derivatives first (square or cube both sides).
If a derivative appears inside sin, cos, log, $e^{()}$ and so on, as in $sin (\frac{d y}{d x})$ or $e^{y^{'}}$ , the equation is not a polynomial in derivatives and its degree is not defined.
Functions of y or x alone (like sin y or $e^{x}$ ) do not affect the degree.

Worked example

Find the order and degree of $[1 + (\frac{d y}{d x})^{2}]^{3/2} = 5 \frac{d ^{2} y}{d x ^{2}}$ . [1 mark]

Try it first, then show the solutionSolution

Square both sides: $[1 + (\frac{d y}{d x})^{2}]^{3} = 25 (\frac{d ^{2} y}{d x ^{2}})^{2}$ , a polynomial in the derivatives.
Highest derivative $\frac{d ^{2} y}{d x ^{2}}$ , raised to power 2.

Answer: Order 2, degree 2

General and particular solutions

NCERT section 9.3

A solution is a function y = φ(x) that satisfies the equation when substituted. The general solution contains arbitrary constants, as many as the order of the equation (it represents a family of curves). A particular solution is obtained by giving the constants definite values, usually from an initial condition such as y(0) = 2.

To verify that a function is a solution, differentiate it the required number of times and substitute into the equation; the two sides must agree.

Worked example

Verify that $y = A e^{2 x} + B e^{- x}$ is a solution of $\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0$ . [2 marks]

Try it first, then show the solutionSolution

$y^{'} = 2 A e^{2 x} - B e^{- x}$ , $y^{''} = 4 A e^{2 x} + B e^{- x}$
$y^{''} - y^{'} - 2 y = (4 - 2 - 2) A e^{2 x} + (1 + 1 - 2) B e^{- x} = 0$

Answer: Verified; it has two arbitrary constants, matching order 2, so it is the general solution.

Variables separable method

NCERT section 9.4.1

If $\frac{d y}{d x} = g (x) h (y)$ , separate the variables and integrate both sides:

$\int \frac{d y}{h ( y )} = \int g (x) d x + C$

Add just one constant. For a particular solution, substitute the given condition to find C. Useful log tidy-ups: $lo g ∣ y ∣ = lo g ∣ sec x ∣ + C$ can be written $y = A sec x$ .

Worked example

Find the particular solution of $\frac{d y}{d x} = y tan x$ , given y = 2 when x = 0. [3 marks]

Try it first, then show the solutionSolution

$\frac{d y}{y} = tan x d x \Rightarrow lo g ∣ y ∣ = lo g ∣ sec x ∣ + C$
So $y = A sec x$ (general solution).
x = 0, y = 2: 2 = A(1), so A = 2.

Answer: $y = 2 sec x$

Homogeneous functions and homogeneous equations

NCERT section 9.4.2

F(x, y) is a homogeneous function of degree n if $F (λ x, λ y) = λ^{n} F (x, y)$ for every non-zero λ. Examples: $x^{2} + x y$ (degree 2), $\frac{x ^{2} + y ^{2}}{x y}$ (degree 0); $x^{2} + y$ is not homogeneous.

The equation $\frac{d y}{d x} = F (x, y)$ is a homogeneous differential equation if F is homogeneous of degree zero. Then F can be written as a function of $\frac{y}{x}$ alone: $\frac{d y}{d x} = g (\frac{y}{x})$ . Similarly $\frac{d x}{d y} = h (\frac{x}{y})$ is homogeneous.

Worked example

Show that $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$ is homogeneous. [1 mark]

Try it first, then show the solutionSolution

$F (λ x, λ y) = \frac{λ ^{2} ( x ^{2} + y ^{2} )}{λ ^{2} x y} = λ^{0} F (x, y)$ , so F is homogeneous of degree 0.

Answer: Homogeneous; $F = \frac{1 + ( y / x ) ^{2}}{y / x}$ .

Solving homogeneous equations: y = vx

NCERT section 9.4.2

For $\frac{d y}{d x} = g (\frac{y}{x})$ , put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . The equation becomes

$v + x \frac{d v}{d x} = g (v) \Rightarrow \frac{d v}{g ( v ) - v} = \frac{d x}{x}$

which is variables separable. Integrate, then replace v by $\frac{y}{x}$ . If the equation is $\frac{d x}{d y} = h (\frac{x}{y})$ (common when $e^{x / y}$ appears), put $x = v y$ instead.

Worked example

Solve $(x^{2} + y^{2}) d x = 2 x y d y$ . [3 marks]

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y}$ , homogeneous. Put y = vx: $v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{2 v}$
$x \frac{d v}{d x} = \frac{1 - v ^{2}}{2 v} \Rightarrow \frac{2 v d v}{1 - v ^{2}} = \frac{d x}{x}$
$- lo g ∣1 - v^{2} ∣ = lo g ∣ x ∣ + C \Rightarrow x (1 - v^{2}) = A \Rightarrow x^{2} - y^{2} = A x$

Answer: $x^{2} - y^{2} = A x$ (A an arbitrary constant)

Linear differential equations dy/dx + Py = Q

NCERT section 9.4.3

A first-order linear equation has the form $\frac{d y}{d x} + P y = Q$ , where P and Q are functions of x (or constants). Method:

1. Write it in this exact form (divide by the coefficient of $\frac{d y}{d x}$ ).
2. Find the integrating factor $IF = e^{\int P d x}$ (no constant needed). Simplify with $e^{l o g f} = f$ .
3. The solution is $y \cdot IF = \int (Q \cdot IF) d x + C$ .

Multiplying by the IF turns the left side into $\frac{d}{d x} (y \cdot IF)$ , which is why the method works.

Worked example

Solve $x \frac{d y}{d x} + 2 y = x^{2}$ , x > 0. [3 marks]

Try it first, then show the solutionSolution

$\frac{d y}{d x} + \frac{2}{x} y = x$ : P = $\frac{2}{x}$ , Q = x.
IF $= e^{\int \frac{2}{x} d x} = e^{2 l o g x} = x^{2}$
$y x^{2} = \int x \cdot x^{2} d x = \frac{x ^{4}}{4} + C$

Answer: $y = \frac{x ^{2}}{4} + \frac{C}{x ^{2}}$

Linear equations in x: dx/dy + P₁x = Q₁

NCERT section 9.4.3

Sometimes the equation is linear in x rather than y, typically when y appears with a high power or inside a function but x appears only to the first power. Rewrite it as

$\frac{d x}{d y} + P_{1} x = Q_{1}$ , with $P_{1}$ , $Q_{1}$ functions of y

Then IF $= e^{\int P_{1} d y}$ and the solution is $x \cdot IF = \int (Q_{1} \cdot IF) d y + C$ . Look for this whenever $\frac{d y}{d x}$ is a fraction whose denominator contains x linearly.

Worked example

Solve $(x + 2 y^{2}) \frac{d y}{d x} = y$ , y > 0. [3 marks]

Try it first, then show the solutionSolution

$\frac{d x}{d y} = \frac{x + 2 y ^{2}}{y} \Rightarrow \frac{d x}{d y} - \frac{1}{y} x = 2 y$
IF $= e^{- \int \frac{d y}{y}} = e^{- l o g y} = \frac{1}{y}$
$\frac{x}{y} = \int 2 y \cdot \frac{1}{y} d y = 2 y + C$

Answer: $x = 2 y^{2} + C y$

Choosing the method

NCERT section 9.4

Before solving, write the equation as $\frac{d y}{d x} = \dots$ and test in this order:

1. Separable? Can the right side be written as (function of x) × (function of y)? Then separate.
2. Homogeneous? Is every term of the same total degree, so that it is a function of $\frac{y}{x}$ ? Then put y = vx.
3. Linear in y? Are y and $\frac{d y}{d x}$ only to the first power and not multiplied together? Then use $e^{\int P d x}$ .
4. Linear in x? Flip to $\frac{d x}{d y}$ and check the same for x.

Some equations fit two methods; either works, but the simpler one saves time.

Worked example

Name the method for (a) $\frac{d y}{d x} = e^{x - y}$ , (b) $\frac{d y}{d x} = \frac{x + y}{x}$ , (c) $\frac{d y}{d x} + y cot x = sin x$ . [1 mark each]

Try it first, then show the solutionSolution

(a) $e^{x - y} = e^{x} \cdot e^{- y}$ : variables separable.
(b) $1 + \frac{y}{x}$ : homogeneous (also linear: $\frac{d y}{d x} - \frac{y}{x} = 1$ ).
(c) Linear with P = cot x, Q = sin x.

Answer: (a) separable (b) homogeneous or linear (c) linear

Maths in real life: Differential Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Exponential growth (separable)

Suppose a lab culture of bacteria grows at a rate proportional to the number present, and a biologist counts 1000 bacteria at the start and 2000 after 3 hours.

Suppose dN/dt = kN with N(0) = 1000 and N(3) = 2000. How many bacteria will there be after 9 hours?

Try it first, then show the solutionSolution

$\frac{d N}{N} = k d t \Rightarrow N = 1000 e^{k t}$
$e^{3 k} = 2$
$N (9) = 1000 (e^{3 k})^{3} = 1000 \times 8 = 8000$

Answer: 8000 bacteria (the count doubles every 3 hours).

Memory hook: Rate proportional to amount means exponential.

Newton's law of cooling

Imagine a cup of tea at 85 °C left in a room at 25 °C, cooling at a rate proportional to the difference between its temperature and the room temperature.

Suppose dT/dt = −k(T − 25), T(0) = 85 and T(5) = 65 (minutes). Find the temperature after 10 minutes.

Try it first, then show the solutionSolution

$\frac{d T}{T - 25} = - k d t \Rightarrow T - 25 = 60 e^{- k t}$
At t = 5: $40 = 60 e^{- 5 k} \Rightarrow e^{- 5 k} = \frac{2}{3}$
At t = 10: $T - 25 = 60 (\frac{2}{3})^{2} = \frac{80}{3}$

Answer: About 51.7 °C.

Memory hook: The gap to room temperature shrinks by the same factor in equal times.

Continuous compounding

Suppose a bank pays interest compounded continuously at 8% a year, so a deposit A grows at the rate dA/dt = 0.08A, and a student deposits ₹10,000.

Find the amount after 5 years.

Try it first, then show the solutionSolution

$A = 10000 e^{0.08 t}$
$A (5) = 10000 e^{0.4} \approx 10000 \times 1.4918$

Answer: About ₹14,918.

Memory hook: Money growing in proportion to itself behaves like a bacteria culture.

Topics in this chapter

Differential Equations formula sheet

Order and degree: $order = highest derivative; degree = its power (polynomial in derivatives only)$
Constants in the general solution: $number of arbitrary constants = order$
Variables separable: $\frac{d y}{d x} = g (x) h (y) \Rightarrow \int \frac{d y}{h ( y )} = \int g (x) d x + C$
Homogeneous function of degree n: $F (λ x, λ y) = λ^{n} F (x, y)$
Homogeneous equation substitution: $y = v x, \frac{d y}{d x} = v + x \frac{d v}{d x}$
Alternative substitution: $x = v y, \frac{d x}{d y} = v + y \frac{d v}{d y}$
Linear in y: $\frac{d y}{d x} + P y = Q : IF = e^{\int P d x}, y \cdot IF = \int Q \cdot IF d x + C$
Linear in x: $\frac{d x}{d y} + P_{1} x = Q_{1} : IF = e^{\int P_{1} d y}, x \cdot IF = \int Q_{1} \cdot IF d y + C$
Useful simplification: $e^{l o g f (x)} = f (x), e^{- l o g f (x)} = \frac{1}{f ( x )}$
Growth and decay: $\frac{d A}{d t} = k A \Rightarrow A = A_{0} e^{k t}$

How the board asks Differential Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Order and degree	1-mark MCQs, often asking for order + degree, or with a derivative inside sin, e or a root.	Clear radicals first; check whether it is a polynomial in derivatives before naming a degree.
Variables separable	MCQs like dy/dx = e^(x − y); 3-mark particular solutions with a given point.	Separate, integrate both sides, one constant, substitute the condition.
Homogeneous equations	3- and 5-mark questions, often with y/x inside cos, tan or e, and a particular solution; MCQs on spotting a homogeneous function.	Show degree 0, put y = vx (or x = vy), separate, back-substitute.
Linear equations and integrating factor	MCQs asking for the IF; 3- and 5-mark solutions in y or in x, sometimes with a condition.	Standard form, IF = e^∫P, then y × IF = ∫Q × IF.
Word problems and case studies	Rates proportional to the amount or the surface area (evaporation, growth, decay, cooling).	Write the rate law as a differential equation, solve by separation, use the data to find constants.

Tricks to remember Differential Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Order first, degree second
Order = how many dashes on the top derivative; degree = its power, but only after clearing roots and only if no derivative hides inside sin, log or e.
Use it when: Order and degree MCQs.
Same degree, same family
If every term in the numerator and denominator has the same total degree, the equation is homogeneous: put y = vx.
Use it when: Spotting homogeneous equations.
IF = e to the P
Standard form first, then e^∫P; then "y times IF equals integral of Q times IF".
Use it when: Linear equations.
Flip when x is lonely
If x appears only to the first power and y appears awkwardly, flip to dx/dy and treat x as the unknown.
Use it when: Linear in x.
Constants match order
Order 1 → one constant; order 2 → two constants; a particular solution has none.
Use it when: General versus particular solution questions.

Differential Equations: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Order and degree together · 1 mark · MCQBoard level

The order and degree of $(\frac{d ^{2} y}{d x ^{2}})^{3} + (\frac{d y}{d x})^{4} + y = 0$ are respectively:
(a) 2, 3
(b) 2, 4
(c) 3, 2
(d) 4, 2

Try it first, then show the solutionSolution

Highest derivative $\frac{d ^{2} y}{d x ^{2}}$ (order 2), raised to power 3 (degree 3).

Answer: (a) 2, 3

Example 2: When degree is not defined · 1 mark · Assertion-reasonBoard level

Assertion (A): The degree of $\frac{d ^{2} y}{d x ^{2}} + sin (\frac{d y}{d x}) = 0$ is not defined.
Reason (R): The degree of a differential equation is defined only when it is a polynomial equation in its derivatives.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$sin (\frac{d y}{d x})$ is not a polynomial in $\frac{d y}{d x}$ , which is exactly why the degree is not defined.

Answer: (a)

Example 3: Separable with exponentials · 2 marks · Very short answerHarder board level

Find the general solution of $\frac{d y}{d x} = e^{x + y}$ . [2 marks]

Try it first, then show the solutionSolution

$e^{- y} d y = e^{x} d x$
$- e^{- y} = e^{x} + C$ , i.e. $e^{x} + e^{- y} = K$

Answer: $e^{x} + e^{- y} = K$

Example 4: A separable particular solution · 3 marks · Short answerHarder board level

Find the particular solution of $(1 + e^{2 x}) d y = (1 + y^{2}) e^{x} d x$ , given y = 0 when x = 0. [3 marks]

Try it first, then show the solutionSolution

$\frac{d y}{1 + y ^{2}} = \frac{e ^{x} d x}{1 + e ^{2 x}}$ ; put $t = e^{x}$ on the right: $tan^{- 1} y = tan^{- 1} (e^{x}) + C$
x = 0, y = 0: $0 = \frac{π}{4} + C$ , so $C = - \frac{π}{4}$
$tan^{- 1} y = tan^{- 1} (e^{x}) - \frac{π}{4}$ , which simplifies to $y = \frac{e ^{x} - 1}{e ^{x} + 1}$

Answer: $y = \frac{e ^{x} - 1}{e ^{x} + 1}$

Example 5: Homogeneous with tan(y/x) · 3 marks · Short answerHarder board level

Solve $x \frac{d y}{d x} = y - x tan (\frac{y}{x})$ . [3 marks]

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y}{x} - tan \frac{y}{x}$ , homogeneous. Put y = vx: $v + x \frac{d v}{d x} = v - tan v$
$cot v d v = - \frac{d x}{x} \Rightarrow lo g ∣ sin v ∣ = - lo g ∣ x ∣ + lo g ∣ C ∣$
$x sin v = C$

Answer: $x sin (\frac{y}{x}) = C$

Example 6: Linear equation with a condition · 5 marks · Long answerHarder board level

Find the particular solution of $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , given y = 1 when x = 0. [5 marks]

Try it first, then show the solutionSolution

Linear with P = tan x: IF $= e^{\int t a n x d x} = e^{l o g s e c x} = sec x$
$y sec x = \int (2 x + x^{2} tan x) sec x d x = \int (2 x sec x + x^{2} sec x tan x) d x$
The integrand is $\frac{d}{d x} (x^{2} sec x)$ , so $y sec x = x^{2} sec x + C$
$y = x^{2} + C cos x$ ; at x = 0: 1 = 0 + C, so C = 1
$y = x^{2} + cos x$

Answer: $y = x^{2} + cos x$

Example 7: Homogeneous equation leading to a perfect square · 5 marks · Long answerHOTS / competency

Find the particular solution of $(x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0$ , given y = 0 when x = 1 (x > 0). [5 marks]

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{x ^{2} + 3 x y + y ^{2}}{x ^{2}} = 1 + 3 \frac{y}{x} + (\frac{y}{x})^{2}$ , homogeneous of degree 0. Put y = vx: $v + x \frac{d v}{d x} = 1 + 3 v + v^{2}$
$x \frac{d v}{d x} = 1 + 2 v + v^{2} = (1 + v)^{2}$ , so $\frac{d v}{( 1 + v ) ^{2}} = \frac{d x}{x}$
Integrate: $- \frac{1}{1 + v} = lo g x + C$
At x = 1, y = 0 (so v = 0): $- 1 = 0 + C$ , C = −1. Hence $\frac{1}{1 + v} = 1 - lo g x$ , i.e. $\frac{x}{x + y} = 1 - lo g x$ .
Solve for y: $x + y = \frac{x}{1 - l o g x}$ , so $y = \frac{x l o g x}{1 - l o g x}$ (for 0 < x < e).

Answer: $\frac{x}{x + y} = 1 - lo g x$ , i.e. $y = \frac{x l o g x}{1 - l o g x}$

Example 8: An evaporating raindrop · 3 marks · Short answerHOTS / competency

A spherical raindrop evaporates at a rate proportional to its surface area. Show that its radius decreases at a constant rate. If the radius falls from 3 mm to 2 mm in 1 hour, when will the drop disappear? [3 marks]

Try it first, then show the solutionSolution

$\frac{d V}{d t} = - k (4 π r^{2})$ and $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ , so $\frac{d r}{d t} = - k$ , a constant.
Then r = 3 − kt (mm, hours); r(1) = 2 gives k = 1, so r = 3 − t.
r = 0 when t = 3.

Answer: The radius decreases at 1 mm per hour; the drop disappears after 3 hours.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

A medicine in the bloodstream

Suppose a doctor gives a patient 200 mg of a medicine, and the body removes it at a rate proportional to the amount A (mg) present: dA/dt = −kA, with t in hours. A blood test shows 100 mg left after 4 hours.

(i) · 1 mark
Write the order and degree of the differential equation.
Answer
Answer: Order 1, degree 1
Only dA/dt appears, to the first power.
(ii) · 1 mark
Solve the equation with A(0) = 200.
Answer
Answer: $A = 200 e^{- k t}$
$\frac{d A}{A} = - k d t \Rightarrow lo g A = - k t + C$ ; A(0) = 200.
(iii) · 2 marks
Find k and the amount left after 6 hours.
Answer
Answer: $k = \frac{l o g 2}{4}$ ; $502 \approx 70.7$ mg
$100 = 200 e^{- 4 k} \Rightarrow e^{- 4 k} = \frac{1}{2}$ . $A (6) = 200 (\frac{1}{2})^{6/4} = 200 \cdot 2^{- 3/2} = \frac{200}{2 2}$ .

A homogeneous equation in class

Suppose a teacher writes the equation $x^{2} \frac{d y}{d x} = x y + y^{2}$ on the board and asks students to solve it step by step.

(i) · 1 mark
Show that the equation is homogeneous.
Answer
Answer: $F (x, y) = \frac{x y + y ^{2}}{x ^{2}}$ satisfies $F (λ x, λ y) = F (x, y)$
Numerator and denominator are both of degree 2, so F has degree 0.
(ii) · 1 mark
After putting y = vx, what separable equation do you get?
Answer
Answer: $\frac{d v}{v ^{2}} = \frac{d x}{x}$
$v + x \frac{d v}{d x} = v + v^{2} \Rightarrow x \frac{d v}{d x} = v^{2}$ .
(iii) · 2 marks
Find the solution passing through (1, 1).
Answer
Answer: $y = \frac{x}{1 - l o g ∣ x ∣}$
$- \frac{1}{v} = lo g ∣ x ∣ + C \Rightarrow - \frac{x}{y} = lo g ∣ x ∣ + C$ ; at (1, 1), C = −1, so $\frac{x}{y} = 1 - lo g ∣ x ∣$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the general solution of $\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$ . [2 marks] (2 marks)

Step	Marks
$\frac{1 - c o s x}{1 + c o s x} = tan^{2} \frac{x}{2} = sec^{2} \frac{x}{2} - 1$	1
$y = 2 tan \frac{x}{2} - x + C$	1

Solve $\frac{d y}{d x} + 3 y = e^{- 2 x}$ . [3 marks] (3 marks)

Step	Marks
Linear with P = 3, Q = $e^{- 2 x}$ ; IF $= e^{3 x}$	1
$y e^{3 x} = \int e^{- 2 x} e^{3 x} d x = \int e^{x} d x = e^{x} + C$	1
$y = e^{- 2 x} + C e^{- 3 x}$	1

Find the particular solution of $(x^{2} + x y) d y = (x^{2} + y^{2}) d x$ , given y = 0 when x = 1. [5 marks] (5 marks)

Step	Marks
$\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x ^{2} + x y}$ is homogeneous (degree 0). Put y = vx: $v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{1 + v}$	1
$x \frac{d v}{d x} = \frac{1 + v ^{2} - v - v ^{2}}{1 + v} = \frac{1 - v}{1 + v} \Rightarrow \frac{1 + v}{1 - v} d v = \frac{d x}{x}$	1
$\frac{1 + v}{1 - v} = - 1 + \frac{2}{1 - v}$ , so $- v - 2 lo g ∣1 - v ∣ = lo g ∣ x ∣ + C$	1
Back-substitute: $- \frac{y}{x} - 2 lo g \frac{x - y}{x} = lo g ∣ x ∣ + C$ ; at (1, 0): C = 0	1
$lo g \frac{( x - y ) ^{2}}{∣ x ∣} = - \frac{y}{x}$ , i.e. $(x - y)^{2} = x e^{- y / x}$ (for x > 0)	1

Where students lose marks in Differential Equations

The slips examiners see most often, and what to write instead.

Mistake: Giving a degree when it is not defined
Answering "degree 1" for $y^{''} + sin (y^{'}) = 0$ .
Do this: If any derivative is inside a non-polynomial function, write "degree not defined".
Usually costs: 1 mark.
Mistake: Not clearing a fractional power before finding degree
Reading $[1 + (y^{'})^{2}]^{3/2} = y^{''}$ as degree 1.
Do this: Square both sides first; the degree is then 2.
Usually costs: 1 mark.
Mistake: Wrong standard form for a linear equation
Finding P without dividing by the coefficient of dy/dx.
Wrong working
$x \frac{d y}{d x} + 2 y = x^{2}$ : P = 2, IF = $e^{2 x}$
Right working
$\frac{d y}{d x} + \frac{2}{x} y = x$ : P = $\frac{2}{x}$ , IF = $x^{2}$
Do this: Make the coefficient of dy/dx equal to 1 before reading P and Q.
Usually costs: Most of the marks.
Mistake: Forgetting to replace v by y/x
Stopping at an answer in v and x.
Do this: Always back-substitute v = y/x (or x/y) and simplify.
Usually costs: 1 mark.
Mistake: Not finding the constant for a particular solution
Leaving C in the answer when a condition is given.
Do this: Substitute the given point, find C and write the final particular solution.
Usually costs: 1 mark.

NCERT exercise guide: Differential Equations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 9.1 (order and degree)

What it tests: Order and degree (if defined) of a variety of differential equations, with MCQs.

How to approach it: Find the highest derivative; check for a polynomial in derivatives; clear radicals.

Same kind as Exercise 9.1

Find the order and degree of $y^{''} + (y^{'})^{2} + sin y = 0$ .

Show the solutionSolution

Highest derivative y″, so order 2.
It is a polynomial in y′ and y″ (sin y involves only y), and y″ appears to power 1.

Answer: Order 2, degree 1

Exercise 9.2 (verifying solutions)

What it tests: Checking that a given explicit or implicit function satisfies a differential equation, and counting arbitrary constants.

How to approach it: Differentiate as often as the order and substitute.

Same kind as Exercise 9.2

Verify that $y = x^{2} + 2 x + C$ is a solution of $y^{'} - 2 x - 2 = 0$ .

Show the solutionSolution

$y^{'} = 2 x + 2$
$y^{'} - 2 x - 2 = 0$ ✓

Answer: Verified

Exercise 9.3 (variables separable)

What it tests: General and particular solutions of separable equations, and word problems on growth, interest and curves with a given slope.

How to approach it: Separate, integrate, one constant; use the condition.

Same kind as Exercise 9.3

Solve $\frac{d y}{d x} = (1 + y^{2}) e^{x}$ .

Show the solutionSolution

$\frac{d y}{1 + y ^{2}} = e^{x} d x$
$tan^{- 1} y = e^{x} + C$

Answer: $tan^{- 1} y = e^{x} + C$

Exercise 9.4 (homogeneous equations)

What it tests: Showing equations are homogeneous and solving them, including particular solutions and x = vy forms.

How to approach it: Degree-0 test; y = vx; separate; back-substitute.

Same kind as Exercise 9.4

Solve $(x + y) d x - x d y = 0$ .

Show the solutionSolution

$\frac{d y}{d x} = 1 + \frac{y}{x}$ ; put y = vx: $v + x v^{'} = 1 + v \Rightarrow d v = \frac{d x}{x}$
$v = lo g ∣ x ∣ + C$

Answer: $y = x lo g ∣ x ∣ + C x$

Exercise 9.5 (linear equations)

What it tests: Linear equations in y and in x, integrating factors, particular solutions and curves defined by slope conditions.

How to approach it: Standard form, IF, integrate Q × IF (often by parts).

Same kind as Exercise 9.5

Solve $\frac{d y}{d x} + y = cos x$ .

Show the solutionSolution

IF $= e^{x}$ ; $y e^{x} = \int e^{x} cos x d x = \frac{e ^{x} ( s i n x + c o s x )}{2} + C$

Answer: $y = \frac{s i n x + c o s x}{2} + C e^{- x}$

Miscellaneous Exercise on Chapter 9

What it tests: Order and degree, verifying solutions, and mixed equations needing any of the three methods, with particular solutions.

How to approach it: Decide the type first; then follow that method.

Same kind as the Miscellaneous Exercise

Find the particular solution of $cos y d x + (1 + e^{- x}) sin y d y = 0$ , given $y = \frac{π}{3}$ when x = 0.

Show the solutionSolution

Divide by $(1 + e^{- x}) cos y$ : $\frac{e ^{x}}{e ^{x} + 1} d x + tan y d y = 0$
$lo g (e^{x} + 1) + lo g ∣ sec y ∣ = lo g A \Rightarrow (e^{x} + 1) sec y = A$
At x = 0, $y = \frac{π}{3}$ : A = 2 × 2 = 4

Answer: $(e^{x} + 1) sec y = 4$

Differential Equations: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.4 marks per paper; 24 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Differential Equations

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 65/1/1, Q11 · 1 mark · MCQ

The integrating factor of differential equation $R \frac{d x}{d y} + P x = Q$ where P, Q, R are functions of y is

(A) $e^{\int \frac{P}{Q} d y}$
(B) $e^{\int P d y}$
(C) $e^{\int \frac{P}{R} d y}$
(D) $e^{\int \frac{P}{R} d x}$

Try it first, then show our solutionSolution

Divide by R: $\frac{d x}{d y} + \frac{P}{R} x = \frac{Q}{R}$ , linear in $x$ . IF $= e^{\int \frac{P}{R} d y}$ . Option (C).
1 mark

Answer: $e^{\int \frac{P}{R} d y}$ , option (C)

Where the marks are

Writes standard form dividing by R and IF w.r.t. y (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q29(a) · 3 marks · Short answer

Find the general solution of the following differential equation :
$x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$

Try it first, then show our solutionSolution

Dividing by $x^{2}$ : $\frac{d y}{d x} = 1 + \frac{y}{x} + (\frac{y}{x})^{2}$ , a homogeneous equation.
½ mark
Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
½ mark
$v + x \frac{d v}{d x} = 1 + v + v^{2} \Rightarrow x \frac{d v}{d x} = 1 + v^{2} \Rightarrow \frac{d v}{1 + v ^{2}} = \frac{d x}{x}$
½ mark
Integrating: $\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x} \Rightarrow tan^{- 1} v = lo g ∣ x ∣ + C$
1 mark
Replacing $v = \frac{y}{x}$ : $tan^{- 1} (\frac{y}{x}) = lo g ∣ x ∣ + C$
½ mark

Answer: $tan^{- 1} (\frac{y}{x}) = lo g ∣ x ∣ + C$

Where the marks are

Writes equation in homogeneous form (½ mark)
Substitutes $y = v x$ (½ mark)
Separates variables $\frac{d v}{1 + v ^{2}} = \frac{d x}{x}$ (½ mark)
Integrates to $tan^{- 1} v = lo g ∣ x ∣ + C$ (1 mark)
Back-substitutes $v = \frac{y}{x}$ (½ mark)

See it in the full 2026 board paper, set 65/1/1

2025, set 65/7/1, Q34 · 5 marks · Long answer

Solve the differential equation $\frac{d y}{d x} = cos x - 2 y$ .

Try it first, then show our solutionSolution

Rewrite the equation in linear form: $\frac{d y}{d x} + 2 y = cos x$ . This is of the type $\frac{d y}{d x} + P y = Q$ with $P = 2$ and $Q = cos x$ .
½ mark
Integrating factor: $I.F. = e^{\int 2 d x} = e^{2 x}$ .
1 mark
The solution is $y \cdot e^{2 x} = \int e^{2 x} cos x d x + C$ .
1½ marks
Let $I_{1} = \int e^{2 x} cos x d x$ . Integrate by parts, taking $cos x$ as the first function:
$I_{1} = cos x \cdot \frac{e ^{2 x}}{2} + \frac{1}{2} \int sin x e^{2 x} d x = \frac{e ^{2 x} c o s x}{2} + \frac{1}{2} [sin x \cdot \frac{e ^{2 x}}{2} - \int cos x \cdot \frac{e ^{2 x}}{2} d x]$
$\Rightarrow I_{1} = \frac{e ^{2 x} c o s x}{2} + \frac{e ^{2 x} s i n x}{4} - \frac{1}{4} I_{1}$
$\Rightarrow \frac{5}{4} I_{1} = \frac{e ^{2 x}}{4} (2 cos x + sin x) \Rightarrow I_{1} = \frac{e ^{2 x}}{5} (2 cos x + sin x)$ .
1½ marks
Hence $y e^{2 x} = \frac{e ^{2 x}}{5} (2 cos x + sin x) + C$ , i.e. the general solution is $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$ .
½ mark

Answer: $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$

Where the marks are

Writes the equation as $\frac{d y}{d x} + 2 y = cos x$ and identifies $P = 2$ , $Q = cos x$ (½ mark)
Finds I.F. $= e^{2 x}$ (1 mark)
Writes the solution form $y e^{2 x} = \int e^{2 x} cos x d x + C$ (1½ marks)
Evaluates $\int e^{2 x} cos x d x = \frac{e ^{2 x}}{5} (2 cos x + sin x)$ by parts (1½ marks)
Writes the general solution $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$ (½ mark)

See it in the full 2025 board paper, set 65/7/1

Every Differential Equations question, by year

For 2027

Sample paper 2026-27: Q11 (1 mark, solved), Q20 (1 mark, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

2026

Set 65/1/1: Q11 (1 mark, solved), Q12 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 65/5/1: Q13 (1 mark, solved), Q14 (1 mark, solved), Q26 (3 marks, solved), Q28 (3 marks, solved)

Sample paper 2025-26: Q10 (1 mark, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

2025

Set 65/1/1: Q4 (1 mark, solved), Q12 (1 mark, solved)

Set 65/7/1: Q34 (5 marks, solved), Q37 (4 marks, solved)

Sample paper 2024-25: Q5 (1 mark, solved), Q14 (1 mark, solved)

2024

Set 65/1/1: Q8 (1 mark, solved), Q14 (1 mark, solved), Q29 (3 marks, solved)

Set 65/5/1: Q14 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Sample paper 2023-24: Q6 (1 mark, solved), Q15 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved), Q31 (3 marks, solved)

2023

Set 65/1/1: Q7 (1 mark, solved), Q15 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 65/5/1: Q12 (1 mark, solved), Q38 (4 marks, solved)

Practice questions on Differential Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The order of $\frac{d ^{3} y}{d x ^{3}} + x (\frac{d y}{d x})^{5} = sin y$ is:
(a) 1
(b) 3
(c) 5
(d) 15
Hint
Look for the highest derivative, not the highest power.
Answer
(b) 3
Q2 · 1 mark · MCQBoard level
The integrating factor of $\frac{d y}{d x} + \frac{y}{x} = x^{2}$ (x > 0) is:
(a) $lo g x$
(b) $e^{x}$
(c) x
(d) $\frac{1}{x}$
Hint
$e^{\int \frac{d x}{x}}$
Answer
(c) x
Q3 · 1 mark · MCQBoard level
The number of arbitrary constants in the particular solution of a second-order differential equation is:
(a) 0
(b) 1
(c) 2
(d) 3
Hint
Particular means the constants have been fixed.
Answer
(a) 0
Q4 · 1 mark · MCQBoard level
The general solution of $y d x + x d y = 0$ is:
(a) x + y = C
(b) xy = C
(c) $x^{2} + y^{2} = C$
(d) $\frac{y}{x} = C$
Hint
$y d x + x d y = d (x y)$ .
Answer
(b) xy = C
Q5 · 2 marks · Very short answerHarder board level
Solve $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ .
Hint
Take $e^{- y}$ common.
Answer
$e^{y} d y = (e^{x} + x^{2}) d x \Rightarrow e^{y} = e^{x} + \frac{x ^{3}}{3} + C$
Q6 · 3 marks · Short answerHarder board level
Solve $x \frac{d y}{d x} + y = x cos x$ .
Hint
The left side is $\frac{d}{d x} (x y)$ .
Answer
$x y = \int x cos x d x = x sin x + cos x + C$ , so $y = sin x + \frac{c o s x + C}{x}$ .
Q7 · 3 marks · Short answerHarder board level
Solve $(x + 2 y) d x - x d y = 0$ as a homogeneous equation, and check your answer by treating it as a linear equation.
Hint
Put y = vx; for the check, write $\frac{d y}{d x} - \frac{2}{x} y = 1$ .
Answer
$\frac{d y}{d x} = 1 + 2 v$ ; $v + x v^{'} = 1 + 2 v \Rightarrow \frac{d v}{1 + v} = \frac{d x}{x} \Rightarrow 1 + v = C x$ , so $x + y = C x^{2}$ . Check: IF $= e^{- \int \frac{2}{x} d x} = \frac{1}{x ^{2}}$ ; $\frac{y}{x ^{2}} = \int \frac{d x}{x ^{2}} = - \frac{1}{x} + C$ , so $y = C x^{2} - x$ , the same family.
Q8 · 3 marks · Short answerHarder board level
Solve $y \frac{d x}{d y} + 2 x = y^{2}$ , y > 0.
Hint
Linear in x: divide by y.
Answer
$\frac{d x}{d y} + \frac{2}{y} x = y$ ; IF $= y^{2}$ ; $x y^{2} = \int y^{3} d y = \frac{y ^{4}}{4} + C$ .
Q9 · 3 marks · Short answerHarder board level
Find the particular solution of $\frac{d y}{d x} = \frac{2 x ( y + 1 )}{x ^{2} + 1}$ , given y = 1 when x = 0.
Hint
Separate; both sides give logs.
Answer
$lo g ∣ y + 1∣ = lo g (x^{2} + 1) + C \Rightarrow y + 1 = A (x^{2} + 1)$ ; A = 2, so $y = 2 x^{2} + 1$ .
Q10 · 5 marks · Long answerHarder board level
Find the particular solution of $x \frac{d y}{d x} - 2 y = x^{3} e^{x}$ (x > 0), given y = 0 when x = 1.
Hint
Divide by x first; IF $= e^{- \int \frac{2}{x} d x}$ .
Answer
$\frac{d y}{d x} - \frac{2}{x} y = x^{2} e^{x}$ ; IF $= e^{- 2 l o g x} = \frac{1}{x ^{2}}$ ; $\frac{y}{x ^{2}} = \int x^{2} e^{x} \cdot \frac{1}{x ^{2}} d x = e^{x} + C$ , so $y = x^{2} e^{x} + C x^{2}$ . At x = 1: 0 = e + C, C = −e. Particular solution: $y = x^{2} (e^{x} - e)$ .
Q11 · 3 marks · Short answerHOTS / competency
Find the equation of the curve through (0, 1) whose slope at every point (x, y) equals x + y.
Hint
$\frac{d y}{d x} - y = x$ is linear.
Answer
IF $= e^{- x}$ ; $y e^{- x} = \int x e^{- x} d x = - x e^{- x} - e^{- x} + C$ , so $y = - x - 1 + C e^{x}$ ; (0, 1) gives C = 2: $y = 2 e^{x} - x - 1$ .
Q12 · 2 marks · Very short answerHOTS / competency
Find the values of k for which $y = e^{k x}$ is a solution of $\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 0$ .
Hint
Substitute and divide by e^(kx).
Answer
$(k^{2} - 5 k + 6) e^{k x} = 0 \Rightarrow k = 2$ or $k = 3$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the order of a differential equation?Show answer
The order of the highest derivative present.
2.When is the degree not defined?Show answer
When the equation is not a polynomial in its derivatives, e.g. a derivative inside sin, log or an exponential.
3.How many arbitrary constants does the general solution of a third-order equation have?Show answer
Three.
4.Condition for dy/dx = F(x, y) to be homogeneous?Show answer
F(λx, λy) = F(x, y), i.e. F homogeneous of degree 0.
5.Substitution for a homogeneous equation?Show answer
y = vx, with dy/dx = v + x dv/dx.
6.Integrating factor of dy/dx + Py = Q?Show answer
$e^{\int P d x}$
7.IF of dy/dx + y/x = x²?Show answer
x (for x > 0).
8.Solution of dA/dt = kA?Show answer
$A = A_{0} e^{k t}$

Best order to study it

Learn the definitions of order and degree and practise MCQs, including equations where the degree is not defined.
Understand general and particular solutions and practise verifying a given solution.
Master variables separable, including particular solutions and word problems on growth and decay.
Learn homogeneous functions, the degree-0 test and the y = vx substitution, then x = vy.
Learn linear equations in y and in x with the integrating factor.
Finish with mixed practice where you first decide which method fits.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the order and degree of an equation with a fractional power?
Can you say when the degree is not defined?
Can you verify that a given function solves a differential equation?
Can you solve a separable equation and find a particular solution?
Can you prove that a function is homogeneous of degree 0?
Can you solve a homogeneous equation by y = vx and by x = vy?
Can you write a linear equation in standard form and find its IF?
Can you solve dx/dy + P₁x = Q₁?
Can you set up and solve a rate-proportional-to-amount problem?
Can you choose the right method quickly?

Differential Equations: frequently asked questions

Is forming a differential equation in the 2026-27 syllabus?

No. Forming a differential equation whose general solution is given was removed in the NCERT rationalisation. You still need order, degree, general and particular solutions and the three solving methods.

Do I add a constant when finding the integrating factor?

No. Any integrating factor works, so take the constant as zero. The single arbitrary constant appears only when you integrate Q times the IF.

Which equations are linear?

Those in which y and its derivative appear only to the first power and are not multiplied together, written as dy/dx plus P times y equals Q. An equation can also be linear in x if you treat x as the unknown.

How important is this chapter?

It is part of Calculus, which carries 35 of the 80 marks. In our question bank it has 38 questions and appears in every board paper from 2023 to 2026, with several 1-mark order and degree questions and a 3- or 5-mark solving question.

Can I leave the answer with log of y?

Yes, an implicit answer is accepted, but simplify it neatly, for example write log y equals log of A sec x as y equals A sec x.

What if an equation is both separable and linear?

Use whichever method is quicker. Both give the same solution, though the constant may look different.

Based on: CBSE Mathematics (041) syllabus 2026-27; NCERT Mathematics Part II, Textbook for Class XII, Chapter 9 (Reprint 2026-27)