MathsISC Class 12 · Differential Equations

Linear Differential Equations: Integrating Factor Method

Updated 15 October 2026 · Fact-checked

A linear differential equation has the form dy/dx + Py = Q, where P and Q are functions of x only. Multiply by the integrating factor IF = e^(∫P dx); the left side becomes d/dx(y × IF), so y × IF = ∫Q × IF dx + C. If x and y swap roles, use dx/dy + Px = Q.

Follow the arrows: write dy/dx + Py = Q, find and simplify the IF, then integrate; if y is not linear, try the dx/dy form.

Illustration drawn for QuestionBuddy.

Understand Linear Differential Equations

Recognising the form. A first-order differential equation is **linear in $y$ ** if it can be written as
$\frac{d y}{d x} + P (x) y = Q (x),$
with $y$ and $\frac{d y}{d x}$ appearing only to the first power and not multiplied together, and $P$ , $Q$ depending on $x$ alone (they may be constants). Examples: $\frac{d y}{d x} + \frac{2}{x} y = x$ , $\frac{d y}{d x} + y tan x = sec x$ . To bring an equation to this form, divide by the coefficient of $\frac{d y}{d x}$ .

The integrating factor. Multiply the whole equation by $IF = e^{\int P d x}$ . Because $\frac{d}{d x} (e^{\int P d x}) = P e^{\int P d x}$ , the left side is exactly the derivative of a product:
$\frac{d}{d x} (y \cdot IF) = Q \cdot IF .$
Integrating,
$y \cdot IF = \int Q \cdot IF d x + C .$
That is the whole method. No constant is needed inside the integrating factor.

Simplify the integrating factor. Often $\int P d x$ is a logarithm, and $e^{l o g f (x)} = f (x)$ :

$P = \frac{2}{x}$ : $IF = e^{2 l o g x} = x^{2}$ .
$P = - \frac{1}{y}$ (in the $x$ -form): $IF = e^{- l o g y} = \frac{1}{y}$ .
$P = tan x$ : $IF = e^{l o g s e c x} = sec x$ .

The ISC 2025 examiners specifically noted candidates who did not simplify integrating factors of the type $e^{l o g f (x)}$ .

**The other form: $\frac{d x}{d y} + P x = Q$ .** Some equations are not linear in $y$ but are linear in $x$ when $y$ is treated as the independent variable. ISC 2025 set $(x + 5 y^{2}) \frac{d y}{d x} = y$ . As written it is neither separable nor linear in $y$ ; but flipping gives $\frac{d x}{d y} = \frac{x + 5 y ^{2}}{y}$ , that is $\frac{d x}{d y} - \frac{1}{y} x = 5 y$ , linear in $x$ with $P = - \frac{1}{y}$ , $Q = 5 y$ . Then $IF = e^{\int P d y}$ and $x \cdot IF = \int Q \cdot IF d y + C$ . The examiners reported candidates who tried the homogeneous method on it and failed: identify the type first.

Particular solutions. If a condition is given, substitute it into the general solution to find $C$ , and write the final answer. The examiners in 2025 again noted candidates who stopped at the general solution.

Why linear equations matter. Many real processes are linear: a tank with brine flowing in and out, a body falling with air resistance proportional to speed, a capacitor charging through a resistor, money in an account with steady deposits. In each, the rate of change equals a constant input minus something proportional to the current amount, $\frac{d y}{d t} + k y = Q$ , which settles towards the steady value $\frac{Q}{k}$ .

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 dy/dx equal to 1 first.
Integrating factor: $IF = e^{\int P d x}$; Simplify e^{log f} = f.
Solution: $y \cdot IF = \int Q \cdot IF d x + C$; One constant C, added after integrating.
Linear in x: $\frac{d x}{d y} + P x = Q, IF = e^{\int P d y}, x \cdot IF = \int Q \cdot IF d y + C$; P, Q functions of y.
Common integrating factors: $P = \frac{k}{x} \Rightarrow IF = x^{k}; P = tan x \Rightarrow sec x; P = k \Rightarrow e^{k x}$; For x > 0 and suitable intervals.
Steady-state model: $\frac{d y}{d t} + k y = Q \Rightarrow y = \frac{Q}{k} + C e^{- k t}$; Constant k and Q; y approaches Q/k.

How to solve Linear Differential Equations questions

Follow these steps; each usually carries a mark in a 5- or 6-mark question.

1Identify the type. If $y$ and $\frac{d y}{d x}$ appear only to the first power, aim for $\frac{d y}{d x} + P y = Q$ . If that fails, try $\frac{d x}{d y} + P x = Q$ .
2Divide through so the coefficient of the derivative is 1, and write $P$ and $Q$ clearly.
3Find $IF = e^{\int P}$ and simplify any $e^{l o g f}$ to $f$ .
4Write the solution formula: $y \cdot IF = \int Q \cdot IF d x + C$ .
5Evaluate the integral (often by parts or substitution).
6Solve for $y$ (or $x$ ) to get the general solution.
7If a condition is given, substitute it to find $C$ and write 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 for steady deposits

Suppose a savings account grows continuously at 5% a year on its balance, and the owner also pays in money continuously at Rs 1000 a year. The balance $B$ (in rupees) after $t$ years satisfies $\frac{d B}{d t} = 0.05 B + 1000$ , starting from $B (0) = 0$ .

Find $B$ as a function of $t$ and the balance after 10 years.

Try it first, then show the solutionSolution

Write it as $\frac{d B}{d t} - 0.05 B = 1000$ . $P = - 0.05$ , so $IF = e^{- 0.05 t}$ .
$B e^{- 0.05 t} = \int 1000 e^{- 0.05 t} d t = - 20000 e^{- 0.05 t} + C$ .
$B = - 20000 + C e^{0.05 t}$ . From $B (0) = 0$ : $C = 20000$ , so $B = 20000 (e^{0.05 t} - 1)$ .
At $t = 10$ : $B = 20000 (e^{0.5} - 1) \approx 20000 \times 0.6487 \approx 12974$ .

Answer: $B = 20000 (e^{0.05 t} - 1)$ ; after 10 years the balance is about Rs 12 974.

Memory hook: Growth plus steady input: a linear equation, solved with IF = e^{−0.05t}.

Linear equation for charging

Imagine a capacitor in a simple circuit whose charge $q$ (in suitable units) after $t$ seconds satisfies $\frac{d q}{d t} + \frac{q}{2} = 3$ , with $q = 0$ at $t = 0$ .

Find $q (t)$ , the charge after 2 seconds, and the final charge.

Try it first, then show the solutionSolution

$IF = e^{t /2}$ . Then $q e^{t /2} = \int 3 e^{t /2} d t = 6 e^{t /2} + C$ .
$q = 6 + C e^{- t /2}$ ; $q (0) = 0$ gives $C = - 6$ , so $q = 6 (1 - e^{- t /2})$ .
At $t = 2$ : $q = 6 (1 - e^{- 1}) \approx 3.79$ . As $t \to \infty$ , $q \to 6$ .

Answer: $q = 6 (1 - e^{- t /2})$ ; about 3.79 units after 2 s, approaching 6 units.

Memory hook: Input minus leak: the amount climbs towards Q/k and never overshoots.

Tricks to remember Linear Differential Equations

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

IF = e to the integral of P
"If you see P, raise e to its integral."
Use it when: Writing the integrating factor.
y times IF equals integral of Q times IF
Multiply y by IF on the left; multiply Q by IF inside the integral on the right.
Use it when: The solution formula.
e and log cancel
e^{log f} = f: e^{2 log x} = x², e^{−log y} = 1/y.
Use it when: Simplifying the IF, as the 2025 examiners asked.
Stuck? Flip it
If the equation has y² or worse with dy/dx, try dx/dy; it may be linear in x.
Use it when: Equations like (x + 5y²) dy/dx = y.

Linear Differential Equations: solved examples

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

Example 1: An integrating factor · 1 mark · MCQBoard level

The integrating factor of $\frac{d y}{d x} + \frac{2}{x} y = x$ ( $x > 0$ ) is:
(a) $2 lo g x$
(b) $x^{2}$
(c) $e^{2 x}$
(d) $\frac{1}{x ^{2}}$

Try it first, then show the solutionSolution

$e^{\int \frac{2}{x} d x} = e^{2 l o g x} = x^{2}$ .

Answer: (b) $x^{2}$ .

Example 2: Constant P · 3 marks · Short answerBoard level

Solve $\frac{d y}{d x} + y = e^{- x}$ .

Try it first, then show the solutionSolution

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

Answer: $y = (x + C) e^{- x}$ .

Example 3: A particular solution with IF = x² · 5 marks · Long answerHarder board level

Solve $x \frac{d y}{d x} + 2 y = x^{2}$ ( $x > 0$ ), given $y = 1$ when $x = 1$ .

Try it first, then show the solutionSolution

Divide by $x$ : $\frac{d y}{d x} + \frac{2}{x} y = x$ . $P = \frac{2}{x}$ , $Q = x$ . (1 mark)
$IF = e^{2 l o g x} = x^{2}$ . (1 mark)
$y x^{2} = \int x \cdot x^{2} d x = \frac{x ^{4}}{4} + C$ . (1 mark)
$y = \frac{x ^{2}}{4} + \frac{C}{x ^{2}}$ . At $(1, 1)$ : $1 = \frac{1}{4} + C$ , so $C = \frac{3}{4}$ . (1 mark)
Particular solution: $y = \frac{x ^{2}}{4} + \frac{3}{4 x ^{2}}$ . (1 mark)

Answer: $y = \frac{x ^{2}}{4} + \frac{3}{4 x ^{2}}$ .

Example 4: Trigonometric P · 3 marks · Short answerHarder board level

Solve $\frac{d y}{d x} + y cot x = 2 cos x$ , for $0 < x < π$ .

Try it first, then show the solutionSolution

$IF = e^{\int c o t x d x} = e^{l o g s i n x} = sin x$ (as $sin x > 0$ ). (1 mark)
$y sin x = \int 2 sin x cos x d x = \int sin 2 x d x = sin^{2} x + C$ . (1 mark)
$y = sin x + C cosec x$ . (1 mark)

Answer: $y = sin x + C cosec x$ .

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

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

Try it first, then show the solutionSolution

Flip: $\frac{d x}{d y} = \frac{x + 3 y ^{3}}{y} = \frac{x}{y} + 3 y^{2}$ , so $\frac{d x}{d y} - \frac{1}{y} x = 3 y^{2}$ : linear in $x$ . (1 mark)
$P = - \frac{1}{y}$ , $IF = e^{- l o g y} = \frac{1}{y}$ . (1 mark)
$\frac{x}{y} = \int 3 y^{2} \cdot \frac{1}{y} d y = \frac{3 y ^{2}}{2} + C$ . (1 mark)
At $x = 2$ , $y = 1$ : $2 = \frac{3}{2} + C$ , so $C = \frac{1}{2}$ . (1 mark)
$\frac{x}{y} = \frac{3 y ^{2} + 1}{2}$ , that is $2 x = 3 y^{3} + y$ . (1 mark)

Answer: $2 x = 3 y^{3} + y$ . (ISC 2025 set the same type with $5 y^{2}$ .)

Example 6: A brine tank · 5 marks · Case-basedHOTS / competency

Suppose a tank holds 100 litres of water in which 10 kg of salt is dissolved. Brine containing 0.5 kg of salt per litre flows in at 4 litres per minute, and the well-mixed solution flows out at the same rate.
(i) Show that the amount of salt $A$ kg after $t$ minutes satisfies $\frac{d A}{d t} + 0.04 A = 2$ . (1 mark)
(ii) Solve for $A (t)$ . (2 marks)
(iii) Find the amount after 25 minutes. (1 mark)
(iv) What amount does the tank approach in the long run? (1 mark)

Try it first, then show the solutionSolution

(i) Salt in: $4 \times 0.5 = 2$ kg/min. Salt out: $4 \times \frac{A}{100} = 0.04 A$ kg/min. So $\frac{d A}{d t} = 2 - 0.04 A$ .
(ii) $IF = e^{0.04 t}$ ; $A e^{0.04 t} = \int 2 e^{0.04 t} d t = 50 e^{0.04 t} + C$ , so $A = 50 + C e^{- 0.04 t}$ . $A (0) = 10$ gives $C = - 40$ : $A = 50 - 40 e^{- 0.04 t}$ .
(iii) $A (25) = 50 - 40 e^{- 1} \approx 50 - 14.72 = 35.28$ kg.
(iv) As $t \to \infty$ , $A \to 50$ kg (the concentration of the incoming brine times the volume).

Answer: $A = 50 - 40 e^{- 0.04 t}$ ; about 35.28 kg after 25 minutes; approaching 50 kg.

Where students lose marks in Linear Differential Equations

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

Mistake: Using the homogeneous method on a linear equation
The ISC 2025 examiners reported candidates who tried y = vx on (x + 5y²) dy/dx = y.
Do this: Identify the type first: if it is not of degree zero in y/x, test whether dy/dx or dx/dy gives a linear form.
Usually costs: The whole 6-mark question.
Mistake: Not simplifying the integrating factor
Leaving IF = e^{2 log x} or e^{−log y} makes the integral look impossible.
Wrong working
$IF = e^{- l o g y}$ , so $x e^{- l o g y} = \int 5 y e^{- l o g y} d y$ .
Right working
$IF = \frac{1}{y}$ , so $\frac{x}{y} = \int 5 d y = 5 y + C$ .
Do this: Always use e^{log f} = f immediately.
Usually costs: 2 to 3 marks.
Mistake: Forgetting to divide by the coefficient of dy/dx
P and Q must be read from the standard form with coefficient 1.
Do this: Write the standard form on its own line before naming P and Q.
Usually costs: The integrating factor and everything after it.
Mistake: Missing the particular solution
The examiners noted candidates who did not use the given condition.
Do this: If x and y values are given, find C and write the final equation.
Usually costs: 1 to 2 marks.

Exam tips

Name the form explicitly: "this is linear of the form $\frac{d x}{d y} + P x = Q$ ".
State $P$ , $Q$ and the integrating factor on separate lines; each is a marking point.
Write the general solution formula before substituting, so a slip in integration still earns method marks.
Use by-parts results you know (for example $\int x e^{x} d x = e^{x} (x - 1)$ ) without re-deriving if time is short.
Check your answer by substituting it into the original equation, at least mentally.

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 $\frac{d x}{d y} + \frac{x}{y} = y^{2}$ ( $y > 0$ ) is:
(a) $x$
(b) $y$
(c) $\frac{1}{y}$
(d) $e^{y}$
Hint
$e^{\int \frac{1}{y} d y}$ .
Answer
(b) $y$ .
Q2 · 3 marks · Short answerBoard level
Solve $\frac{d y}{d x} - y = 2 x$ .
Hint
$IF = e^{- x}$ ; integrate $2 x e^{- x}$ by parts.
Answer
$y e^{- x} = \int 2 x e^{- x} d x = - 2 e^{- x} (x + 1) + C$ , so $y = - 2 (x + 1) + C e^{x}$ .
Q3 · 3 marks · Short answerHarder board level
Solve $(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x ^{2}}$ .
Hint
The left side is already $\frac{d}{d x} [(1 + x^{2}) y]$ .
Answer
$P = \frac{2 x}{1 + x ^{2}}$ , $IF = 1 + x^{2}$ . $y (1 + x^{2}) = \int \frac{d x}{1 + x ^{2}} = tan^{- 1} x + C$ , so $y = \frac{t a n ^{- 1} x + C}{1 + x ^{2}}$ .
Q4 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} + 2 y = sin x$ .
Hint
Use $\int e^{2 x} sin x d x = \frac{e ^{2 x} ( 2 s i n x - c o s x )}{5}$ .
Answer
$y e^{2 x} = \frac{e ^{2 x} ( 2 s i n x - c o s x )}{5} + C$ , so $y = \frac{2 s i n x - c o s x}{5} + C e^{- 2 x}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d y}{d x} + \frac{y}{x} = x^{2}$ is a linear differential equation.
Reason (R): Its integrating factor is $x$ (for $x > 0$ ).
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
What makes an equation linear?
Answer
(b). Both are true, but the equation is linear because $y$ and $\frac{d y}{d x}$ appear to the first power with coefficients in $x$ ; the integrating factor does not explain that.
Q6 · 3 marks · Short answerHOTS / competency
Solve $x lo g x \frac{d y}{d x} + y = \frac{2}{x} lo g x$ ( $x > 1$ ).
Hint
$P = \frac{1}{x l o g x}$ , so $IF = lo g x$ .
Answer
$y lo g x = \int \frac{2}{x ^{2}} lo g x d x = - \frac{2}{x} (1 + lo g x) + C$ (by parts). So $y lo g x = - \frac{2}{x} (1 + lo g x) + C$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a parachutist falls from rest, and with the parachute open her velocity $v$ (m/s) satisfies $\frac{d v}{d t} + 0.5 v = 9.8$ .
(i) Find the integrating factor. (1 mark)
(ii) Find $v (t)$ . (1 mark)
(iii) Find her velocity after 2 seconds. (1 mark)
(iv) What velocity does she approach? (1 mark)
Hint
$v (0) = 0$ .
Answer
(i) $e^{0.5 t}$ . (ii) $v e^{0.5 t} = 19.6 e^{0.5 t} + C$ , $C = - 19.6$ : $v = 19.6 (1 - e^{- 0.5 t})$ . (iii) $19.6 (1 - e^{- 1}) \approx 12.39$ m/s. (iv) 19.6 m/s, the terminal velocity.

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.Write the standard form of a linear equation in $y$ .Show answer
$\frac{d y}{d x} + P y = Q$ , with $P$ , $Q$ functions of $x$ .
2.What is the integrating factor?Show answer
$e^{\int P d x}$ .
3.Find the IF of $\frac{d y}{d x} - \frac{3}{x} y = x$ .Show answer
$x^{- 3}$ .
4.Write the solution formula.Show answer
$y \cdot IF = \int Q \cdot IF d x + C$ .
5.When should you use $\frac{d x}{d y} + P x = Q$ ?Show answer
When the equation is linear in $x$ (with $y$ as the independent variable) but not in $y$ .
6.What value does $y$ approach if $\frac{d y}{d t} + 2 y = 10$ ?Show answer
5.

Linear Differential Equations: frequently asked questions

How do I recognise a linear differential equation?

After dividing by the coefficient of dy/dx, it should read dy/dx + Py = Q, where P and Q depend on x only. y and dy/dx must appear only to the first power and not multiplied together.

Why does the integrating factor work?

Multiplying by e^(∫P dx) turns the left side into the derivative of y times that factor, by the product rule. The equation can then be integrated in one step.

What if the equation is not linear in y?

Try treating x as the dependent variable. If it can be written as dx/dy + Px = Q, with P and Q functions of y, it is linear in x and the same method works with integrals in y.

Do I need a constant in the integrating factor?

No. Any constant multiple of the integrating factor works, so the constant of integration inside it can be ignored. Only one constant C is needed, in the final solution.

Are linear differential equations in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus includes linear differential equations of both forms, dy/dx + Py = Q and dx/dy + Px = Q, and a 6-mark question in ISC 2025 used the second form.