Match List-I with List-II, List-I (Differential equation), List-II (Integrating factor), (A) dy/dx + 2y = x, (I) e^x, (B) dy/dx + (2/x)y = x^2, (II) e^2x, (C) dy/dx - (2/x)y = x, (III) 1/x^2, (D) dy/dx + y = e^(-x), (IV) x^2, Choose the correct answer from the options given below: Option 1: (A) - (II), (B) - (IV), (C) - (III), (D) - (I) Option 2: (A) - (I), (B) - (II), (C) - (IV), (D) - (III) Option 3: (A) - (III), (B) - (I), (C) - (II), (D) - (IV) Option 4: (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

Answer: (A) - (II), (B) - (IV), (C) - (III), (D) - (I). Solution: For dy/dx + Py = Q, IF = e^( P dx). (A) P = 2 e^2x. (B) P = 2/x x^2. (C) P = -2/x 1/x^2. (D) P = 1 e^x.

Calculus

Differential Equations

Easy

core

Match List-I with List-II

List-I (Differential equation) List-II (Integrating factor)
(A) $\frac{dy}{dx} + 2y = x$ (I) $e^x$
(B) $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x^2$ (II) $e^{2x}$
(C) $\frac{dy}{dx} - \left(\frac{2}{x}\right)y = \sin x$ (III) $\frac{1}{x^2}$
(D) $\frac{dy}{dx} + y = e^{-x}$ (IV) $x^2$

Choose the correct answer from the options given below:

List-I (Differential equation)	List-II (Integrating factor)
(A) $\frac{dy}{dx} + 2y = x$	(I) $e^x$
(B) $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x^2$	(II) $e^{2x}$
(C) $\frac{dy}{dx} - \left(\frac{2}{x}\right)y = \sin x$	(III) $\frac{1}{x^2}$
(D) $\frac{dy}{dx} + y = e^{-x}$	(IV) $x^2$

(A) - (II), (B) - (IV), (C) - (III), (D) - (I)

(A) - (I), (B) - (II), (C) - (IV), (D) - (III)

(A) - (III), (B) - (I), (C) - (II), (D) - (IV)

(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

✅ Correct Option: 1

For $dy/dx + Py = Q$ , IF $= e^{\int P dx}$ . (A) $P = 2 \to e^{2x}$ . (B) $P = 2/x \to x^2$ . (C) $P = -2/x \to 1/x^2$ . (D) $P = 1 \to e^{x}$ .

2025: 29 May Shift 2

The particular solution of the differential equation $\frac{dy}{dx} = 8yx$ when $y = 1$ at $x = 0$

2026: 18 May Shift 1

The solution of the differential equation $x\frac{dy}{dx} = 2x^2 + 1$ ; $x \neq 0$ with $y = 1$ when $x = 1$ , is

2026: 22 May Shift 2

For a given differential equation $e^{dy/dx} = x + 1; x \in (-1,\infty)$ , then which of the following statements are TRUE?

(A) Its general solution is $y = x\log_e(x+1) - x + \log_e(x+1) + C$ ; C is an arbitrary constant.

(B) Its particular solution is $y = (x+1)\log_e(x+1) - x + 5; y(0) = 5$

(D) Its particular solution is $y = (x-1)\log(x+1) - x + 5; y(0) = 5$

Choose the correct answer from the options given below:

2025: 13 May Shift 1

The general solution of the differential equation $\log_e\left(\frac{dy}{dx}\right) = ax + by$ is

2022: 10 Aug Shift 1

The integrating factor of differential equation $x\frac{dy}{dx} + 2y = x^2 \log x$ is

2026: 15 May Shift 1

If $c$ is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I (Differential Equation)		LIST-II (Solution)
A.	$\dfrac{dy}{dx} = 1 + x + y + xy$	I.	$\log_e \vert y \vert = 2x + \log_e (x-1)^2 + c$
B.	$(x-1)\dfrac{dy}{dx} = 2xy$	II.	$y = cx$
C.	$\dfrac{dy}{dx} = 1 - x + y - xy$	III.	$\log_e \vert 1+y \vert = x + \dfrac{x^2}{2} + c$
D.	$x\,dy = y\,dx$	IV.	$\log_e \vert 1+y \vert = x - \dfrac{x^2}{2} + c$

Choose the correct answer from the options given below:

Solution

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