MathsCBSE Class 12 · Differential Equations

Linear Differential Equations and the Integrating Factor (Class 12)

Updated 15 October 2026 · Fact-checked

A linear equation has the form dy/dx + Py = Q, with P and Q functions of x. Multiply by the integrating factor IF = e^∫P dx; the left side becomes d/dx(y·IF), so y·IF = ∫Q·IF dx + C. For dx/dy + P₁x = Q₁, the roles of x and y swap.

Follow the steps down: find IF = e^∫P dx, multiply, and the left side becomes d/dx(y·IF); if x² or y³ blocks you, swap x and y.

Illustration drawn for QuestionBuddy.

Understand Linear Differential Equations

Standard form. A first-order differential equation is linear in y if it can be written as
$\frac{d y}{d x} + P y = Q$ ,
where $P$ and $Q$ are functions of $x$ alone (or constants). The unknown $y$ and its derivative appear only to the first power and are not multiplied together. Examples: $\frac{d y}{d x} + 2 y = e^{- x}$ ; $x \frac{d y}{d x} + 2 y = x^{2}$ (divide by $x$ first: $P = \frac{2}{x}$ , $Q = x$ ).

The integrating factor. Multiply the whole equation by $IF = e^{\int P d x}$ . Because $\frac{d}{d x} (IF) = P \cdot IF$ , the left side becomes the derivative of a product:
$IF \frac{d y}{d x} + P IF y = \frac{d}{d x} (y \cdot IF)$ .
Integrating both sides gives the general solution
$y \cdot IF = \int Q \cdot IF d x + C$ .
Useful simplifications: $e^{l o g x} = x$ , $e^{2 l o g x} = x^{2}$ , $e^{- l o g x} = \frac{1}{x}$ , $e^{l o g s i n x} = sin x$ , $e^{l o g s e c x} = sec x$ . No constant is needed when finding the IF.

Linear in x. Some equations are not linear in $y$ but become linear when $x$ is treated as the unknown: $\frac{d x}{d y} + P_{1} x = Q_{1}$ , with $P_{1}, Q_{1}$ functions of $y$ . Then $IF = e^{\int P_{1} d y}$ and $x \cdot IF = \int Q_{1} \cdot IF d y + C$ . A typical sign: $y^{2}$ or $y^{3}$ appears with $x$ , as in $y d x - (x + 2 y^{3}) d y = 0$ , or the equation is $(x + y + 2) \frac{d y}{d x} = 1$ .

The integration step. $\int Q \cdot IF d x$ often needs substitution or integration by parts (for example $\int x e^{- x} d x$ or $\int e^{x} sin x d x$ , where by parts is applied twice and the original integral reappears).

Applications. Linear equations model a quantity that changes by a proportional amount plus a steady input: a tank receiving salt while mixing, a savings account with continuous interest and regular deposits, a current building up in a circuit. They all have the form $\frac{d Q}{d t} + k Q = R$ and solutions that settle towards $\frac{R}{k}$ .

Both forms ( $\frac{d y}{d x} + P y = Q$ and $\frac{d x}{d y} + P_{1} x = Q_{1}$ ) are in the 2026-27 syllabus (NCERT Section 9.4.3) and are asked every year, often as 1-mark integrating-factor MCQs and 3- or 5-mark solutions.

Linear Differential Equations: formulas to remember

Linear in y: $\frac{d y}{d x} + P y = Q, P, Q functions of x$; Make the coefficient of $\frac{d y}{d x}$ equal to 1 first.
Integrating factor: $IF = e^{\int P d x}$; No constant in this integral.
General solution: $y \cdot IF = \int Q \cdot IF d x + C$; Divide by IF to get y if needed.
Linear in x: $\frac{d x}{d y} + P_{1} x = Q_{1} : x \cdot e^{\int P_{1} d y} = \int Q_{1} e^{\int P_{1} d y} d y + C$; $P_{1}, Q_{1}$ are functions of y.
Useful IF values: $e^{l o g x} = x, e^{- l o g x} = \frac{1}{x}, e^{l o g s e c x} = sec x, e^{l o g s i n x} = sin x$; Simplify the IF before using it.
Constant coefficients: $\frac{d y}{d x} + k y = R \Rightarrow y = \frac{R}{k} + C e^{- k x}$; k and R constants; the solution settles to $\frac{R}{k}$ .

How to solve Linear Differential Equations questions

The standard layout for a linear equation (3 or 5 marks).

1Rewrite in standard form $\frac{d y}{d x} + P y = Q$ (or $\frac{d x}{d y} + P_{1} x = Q_{1}$ ); divide through by the coefficient of the derivative.
2Identify $P$ and $Q$ clearly, with signs.
3Find $IF = e^{\int P d x}$ and simplify it.
4Write $y \cdot IF = \int Q \cdot IF d x + C$ and evaluate the integral (by substitution or parts if needed).
5Simplify, and use any given condition to find C for the particular solution.

Linear Differential Equations in real life

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.

Linear equation with a steady input

Suppose a student's savings account earns interest continuously at 5% a year, and she also deposits money continuously at Rs 12,000 a year, starting from a zero balance. The balance $A$ satisfies $\frac{d A}{d t} = 0.05 A + 12000$ .

Solve for $A (t)$ and find the balance after 10 years.

Try it first, then show the solutionSolution

Standard form: $\frac{d A}{d t} - 0.05 A = 12000$ , so $P = - 0.05$ and $IF = e^{- 0.05 t}$ .
$A e^{- 0.05 t} = \int 12000 e^{- 0.05 t} d t = - 240000 e^{- 0.05 t} + C$ , so $A = - 240000 + C e^{0.05 t}$ .
$A (0) = 0 \Rightarrow C = 240000$ : $A = 240000 (e^{0.05 t} - 1)$ .
$A (10) = 240000 (e^{0.5} - 1) \approx 240000 \times 0.6487 \approx$ Rs 1,55,700.

Answer: $A = 240000 (e^{0.05 t} - 1)$ ; about Rs 1,55,700 after 10 years.

Memory hook: Growth plus a steady input: a linear equation with an integrating factor.

Approach to a steady value

Imagine the current $i$ (in amperes) in a simple circuit just after switching on satisfies $\frac{d i}{d t} + 2 i = 10$ , with $i = 0$ at $t = 0$ seconds.

Find $i (t)$ , the current after 1 second, and the value it approaches.

Try it first, then show the solutionSolution

$IF = e^{2 t}$ : $i e^{2 t} = \int 10 e^{2 t} d t = 5 e^{2 t} + C$ , so $i = 5 + C e^{- 2 t}$ .
$i (0) = 0 \Rightarrow C = - 5$ : $i = 5 (1 - e^{- 2 t})$ .
$i (1) = 5 (1 - e^{- 2}) \approx 4.32$ A.
As $t$ grows, $e^{- 2 t} \to 0$ , so $i \to 5$ A.

Answer: $i = 5 (1 - e^{- 2 t})$ ; about 4.32 A at 1 s; it approaches 5 A.

Memory hook: $\frac{d Q}{d t} + k Q = R$ settles at $\frac{R}{k}$ .

Tricks to remember Linear Differential Equations

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

One in front, P and Q visible
Make the derivative's coefficient 1, then read P (with y) and Q (alone).
Use it when: Avoiding a wrong integrating factor.
e to the integral of P
IF $= e^{\int P}$ ; then 'y times IF equals integral of Q times IF, plus C'.
Use it when: The solution formula.
e and log cancel
$e^{l o g f (x)} = f (x)$ : always simplify the IF to a plain function.
Use it when: IFs like $x^{2}$ , $sec x$ , $sin x$ .
Swap roles when y is the troublemaker
If y appears in powers or functions next to x, try $\frac{d x}{d y}$ : it may be linear in x.
Use it when: Equations like $(x + y + 2) \frac{d y}{d x} = 1$ .

Linear Differential Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Constant coefficients · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

$P = 2$ , $Q = e^{- x}$ ; $IF = e^{2 x}$ . (1 mark)
$y e^{2 x} = \int e^{- x} e^{2 x} d x = \int e^{x} d x = e^{x} + C$ . (1 mark)
$y = e^{- x} + C e^{- 2 x}$ . (1 mark)

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

Example 2: Dividing by x first · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Standard form: $\frac{d y}{d x} + \frac{2}{x} y = x^{2}$ ; $IF = e^{2 l o g x} = x^{2}$ . (1 mark)
$y x^{2} = \int x^{2} \cdot x^{2} d x = \frac{x ^{5}}{5} + C$ . (1 mark)
$y = \frac{x ^{3}}{5} + \frac{C}{x ^{2}}$ . (1 mark)

Answer: $y = \frac{x ^{3}}{5} + \frac{C}{x ^{2}}$

Example 3: A trigonometric IF and a particular solution · 3 marks · Short answerHarder board level

Solve $\frac{d y}{d x} + y cot x = 2 cos x$ , given $y = 2$ when $x = \frac{π}{2}$ .

Try it first, then show the solutionSolution

$P = cot x$ ; $IF = e^{l o g s i n x} = sin x$ . (1 mark)
$y sin x = \int 2 sin x cos x d x = sin^{2} x + C$ , so $y = sin x + C cosec x$ . (1 mark)
At $x = \frac{π}{2}$ : $2 = 1 + C \Rightarrow C = 1$ ; $y = sin x + cosec x$ . (1 mark)

Answer: $y = sin x + cosec x$

Example 4: Linear in x · 5 marks · Long answerHarder board level

Solve $y d x - (x + 2 y^{3}) d y = 0$ , given $x = 2$ when $y = 1$ .

Try it first, then show the solutionSolution

Rewrite: $\frac{d x}{d y} = \frac{x + 2 y ^{3}}{y}$ , i.e. $\frac{d x}{d y} - \frac{1}{y} x = 2 y^{2}$ : linear in x with $P_{1} = - \frac{1}{y}$ , $Q_{1} = 2 y^{2}$ . (1½ marks)
$IF = e^{- l o g y} = \frac{1}{y}$ (for $y > 0$ ). (1 mark)
$\frac{x}{y} = \int 2 y^{2} \cdot \frac{1}{y} d y = y^{2} + C$ . (1 mark)
$x = y^{3} + C y$ . With $x = 2$ , $y = 1$ : $2 = 1 + C$ , so $C = 1$ . (1 mark)
Particular solution: $x = y^{3} + y$ . (½ mark)

Answer: $x = y^{3} + y$

Example 5: Integration by parts twice inside · 5 marks · Long answerHOTS / competency

Find the particular solution of $\frac{d y}{d x} + y = sin x$ , given $y (0) = 0$ .

Try it first, then show the solutionSolution

$IF = e^{x}$ ; $y e^{x} = \int e^{x} sin x d x$ . (1 mark)
Let $J = \int e^{x} sin x d x$ . By parts: $J = e^{x} sin x - \int e^{x} cos x d x = e^{x} sin x - (e^{x} cos x + \int e^{x} sin x d x)$ . (1½ marks)
So $2 J = e^{x} (sin x - cos x)$ and $J = \frac{e ^{x} ( s i n x - c o s x )}{2}$ . (1 mark)
$y = \frac{s i n x - c o s x}{2} + C e^{- x}$ ; $y (0) = 0 \Rightarrow - \frac{1}{2} + C = 0 \Rightarrow C = \frac{1}{2}$ . (1 mark)
$y = \frac{s i n x - c o s x + e ^{- x}}{2}$ . (½ mark)

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

Example 6: Case study: salt in a mixing tank · 4 marks · Case-basedHOTS / competency

Suppose a 100-litre tank holds brine with 10 kg of dissolved salt. Brine containing 0.2 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at 5 L/min. If $S (t)$ kg is the salt after $t$ minutes, then $\frac{d S}{d t} = 1 - \frac{S}{20}$ .
(i) Explain the two terms on the right. (1 mark)
(ii) Solve for $S (t)$ . (2 marks)
(iii) Find the salt after 20 minutes and the long-run amount. (1 mark)

Try it first, then show the solutionSolution

(i) Inflow: $5 \times 0.2 = 1$ kg/min. Outflow: concentration $\frac{S}{100}$ kg/L times 5 L/min $= \frac{S}{20}$ kg/min. (1 mark)
(ii) $\frac{d S}{d t} + \frac{S}{20} = 1$ ; $IF = e^{t /20}$ ; $S e^{t /20} = 20 e^{t /20} + C$ , so $S = 20 + C e^{- t /20}$ . (1 mark)
$S (0) = 10 \Rightarrow C = - 10$ : $S = 20 - 10 e^{- t /20}$ . (1 mark)
(iii) $S (20) = 20 - \frac{10}{e} \approx 16.3$ kg; in the long run $S \to 20$ kg (the incoming concentration times 100 L). (1 mark)

Answer: (ii) $S = 20 - 10 e^{- t /20}$ (iii) about 16.3 kg; approaches 20 kg.

Where students lose marks in Linear Differential Equations

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

Mistake: Not dividing by the coefficient of dy/dx
P must be read from the form with coefficient 1.
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: Always rewrite in standard form first.
Usually costs: The whole solution.
Mistake: Wrong sign of P
$\frac{d y}{d x} - \frac{y}{x} = x$ has $P = - \frac{1}{x}$ .
Wrong working
$IF = e^{l o g x} = x$ .
Right working
$IF = e^{- l o g x} = \frac{1}{x}$ .
Do this: Copy P with its sign.
Usually costs: 2 to 3 marks.
Mistake: Forgetting C or putting it in the wrong place
The constant belongs with the integral before dividing by IF.
Wrong working
$y = \frac{x ^{2}}{4} + C$ .
Right working
$y x^{2} = \frac{x ^{4}}{4} + C \Rightarrow y = \frac{x ^{2}}{4} + \frac{C}{x ^{2}}$ .
Do this: Add C on the right of $y \cdot IF = \dots$ and then divide everything.
Usually costs: 1 mark.
Mistake: Forcing a linear-in-x equation into y-form
Some equations are linear only in x.
Wrong working
$y d x - (x + 2 y^{3}) d y = 0$ treated as $\frac{d y}{d x} + P y = Q$ .
Right working
$\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ , linear in x.
Do this: If y appears in powers alongside x, try $\frac{d x}{d y}$ .
Usually costs: Most of the marks.

Exam tips

Linear equations were asked in the 2023, 2024, 2025 and 2026 board papers, with 1-mark IF MCQs and 3- or 5-mark solutions; write P, Q and IF on separate lines.
Simplify the IF using $e^{l o g f} = f$ before multiplying.
Show the integral $\int Q \cdot IF d x$ clearly; it is usually where the method marks are.
For particular solutions, find C from the general solution, then write the final answer in a single line.
Remember the second form $\frac{d x}{d y} + P_{1} x = Q_{1}$ ; it is explicitly in the syllabus.

Practice questions on Linear 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 integrating factor of $\frac{d y}{d x} - \frac{y}{x} = x$ ( $x > 0$ ) is:
(a) $x$
(b) $\frac{1}{x}$
(c) $e^{x}$
(d) $lo g x$
Hint
$P = - \frac{1}{x}$ .
Answer
(b) $e^{- l o g x} = \frac{1}{x}$ .
Q2 · 2 marks · Very short answerBoard level
Find the integrating factor of $\frac{d y}{d x} + y tan x = sec x$ .
Hint
$\int tan x d x = lo g sec x$ .
Answer
$IF = e^{\int t a n x d x} = e^{l o g s e c x} = sec x$ .
Q3 · 2 marks · Very short answerBoard level
Solve $\frac{d y}{d x} + 3 y = 6$ .
Hint
$IF = e^{3 x}$ .
Answer
$y e^{3 x} = \int 6 e^{3 x} d x = 2 e^{3 x} + C$ , so $y = 2 + C e^{- 3 x}$ .
Q4 · 3 marks · Short answerHarder board level
Solve $x \frac{d y}{d x} - y = x^{2}$ , $x > 0$ .
Hint
Standard form $\frac{d y}{d x} - \frac{1}{x} y = x$ .
Answer
$IF = \frac{1}{x}$ : $\frac{y}{x} = \int 1 d x = x + C$ , so $y = x^{2} + C x$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The integrating factor of $\frac{d y}{d x} + \frac{1}{x} y = x^{2}$ ( $x > 0$ ) is $x$ .
Reason (R): The integrating factor of $\frac{d y}{d x} + P y = Q$ is $e^{\int P d x}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute $e^{\int \frac{1}{x} d x}$ .
Answer
(a). $e^{l o g x} = x$ , exactly by the rule in R.
Q6 · 5 marks · Long answerHarder board level
Find the particular solution of $\frac{d y}{d x} + y tan x = cos^{2} x$ , given $y (0) = 1$ .
Hint
$IF = sec x$ , and $cos^{2} x sec x = cos x$ .
Answer
$y sec x = \int cos x d x = sin x + C$ , so $y = sin x cos x + C cos x$ . $y (0) = 1 \Rightarrow C = 1$ : $y = cos x (sin x + 1)$ .
Q7 · 3 marks · Short answerHOTS / competency
Solve $(x + y + 2) \frac{d y}{d x} = 1$ , given $x = 0$ when $y = 0$ .
Hint
Write $\frac{d x}{d y} - x = y + 2$ .
Answer
$IF = e^{- y}$ : $x e^{- y} = \int (y + 2) e^{- y} d y = - (y + 2) e^{- y} - e^{- y} + C = - (y + 3) e^{- y} + C$ . So $x = - (y + 3) + C e^{y}$ ; with $(0, 0)$ : $C = 3$ , giving $x = 3 e^{y} - y - 3$ .

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.Standard form of a linear equation in y?Show answer
$\frac{d y}{d x} + P y = Q$ , P and Q functions of x.
2.Integrating factor?Show answer
$e^{\int P d x}$ .
3.General solution?Show answer
$y \cdot IF = \int Q \cdot IF d x + C$ .
4.IF of $\frac{d y}{d x} + \frac{3}{x} y = x$ ?Show answer
$x^{3}$ .
5.IF of $\frac{d y}{d x} + y cot x = 1$ ?Show answer
$sin x$ .
6.Form of a linear equation in x?Show answer
$\frac{d x}{d y} + P_{1} x = Q_{1}$ , $P_{1}, Q_{1}$ functions of y.

Linear Differential Equations: frequently asked questions

How do I recognise a linear differential equation?

After rearranging, it should read dy/dx + Py = Q where P and Q involve only x (or constants). The unknown y and dy/dx appear only to the first power and are not multiplied together or inside other functions.

Why does the integrating factor work?

Multiplying by e^∫P dx makes the left side exactly the derivative of y times the integrating factor, by the product rule. Then a single integration solves the equation.

Do I add a constant when finding the integrating factor?

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

Which NCERT section covers linear equations?

Section 9.4.3 (Linear Differential Equations) of Chapter 9 in NCERT Mathematics Part II for Class 12.

When should I use dx/dy instead of dy/dx?

When the equation becomes linear in x but not in y, for example when y appears in powers or functions next to x, as in y dx − (x + 2y³) dy = 0.