For a given differential equation e^(dy/dx) = x + 1; x (-1,∞), then which of the following statements are TRUE? (A) Its general solution is y = x_e(x+1) - x + _e(x+1) + C; C is an arbitrary constant. (B) Its particular solution is y = (x+1)_e(x+1) - x + 5; y(0) = 5 (C) Its general solution is y = (x-1)(x+1) - x + C; C is an arbitrary constant (D) Its particular solution is y = (x-1)(x+1) - x + 5; y(0) = 5 Choose the correct answer from the options given below: Option 1: (A) and (D) only Option 2: (C) and (D) only Option 3: (A) and (B) only Option 4: (B) and (C) only

Answer: (A) and (B) only. Solution: Taking logarithms, dy/dx=_e(x+1). Integrating by parts, y=(x+1)_e(x+1)-(x+1)+C, which absorbs the constant to y=(x+1)_e(x+1)-x+C. This is exactly statement (A) rewritten. With y(0)=5: 0-0+C=5, so C=5, giving statement (B). Statements (C) and (D) use (x-1) and fail.

Calculus

Differential Equations

Medium

common

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 $y = (x+1)\log_e(x+1) - x + 5; y(0) = 5$

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

(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:

(A) and (D) only

(A) and (B) only

(B) and (C) only

✅ Correct Option: 3

Taking logarithms, $\frac{dy}{dx}=\log_e(x+1)$ .

Integrating by parts, $y=(x+1)\log_e(x+1)-(x+1)+C$ , which absorbs the constant to $y=(x+1)\log_e(x+1)-x+C$ . This is exactly statement (A) rewritten.

With $y(0)=5$ : $0-0+C=5$ , so $C=5$ , giving statement (B). Statements (C) and (D) use $(x-1)$ and fail.

2026: 22 May Shift 1

Integrating factor of the differential equation $x(x-1)\frac{dy}{dx}-(x-2)y=x^3(2x-1), x>1$ is:

2022: 30 Aug Shift 1

The integrating factor of the differential equation $\cos x \frac{dy}{dx} + y\sin x = 1$ is

2026: 26 May Shift 2

Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Order and Degree
A.	$\left(\dfrac{dy}{dx}\right)^2 + 3y\left(\dfrac{d^2y}{dx^2}\right) = 0$	I.	$1,1$
B.	$\sqrt{1-\left(\dfrac{dy}{dx}\right)^2} = a\left(\dfrac{d^2y}{dx^2}\right)^{1/3}$	II.	$1,2$
C.	$y = \dfrac{dy}{dx} + \dfrac{5}{(dy/dx)}$	III.	$2,1$
D.	$\sqrt{1-y^2}\,dx + \sqrt{1-x^2}\,dy = 0$	IV.	$2,2$

Choose the correct answer from the options given below:

2022: 10 Aug Shift 1

General solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = 0$ is

A. $\tan^{-1} x + \tan^{-1} y = C$

B. $\sin^{-1} x - \cos^{-1} y = C$

C. $x\sqrt{1 - y^2} - y\sqrt{1 - x^2} = C$

D. $\sin^{-1} x + \sin^{-1} y = C$

E. $\cos^{-1} x + \cos^{-1} y = C$

(where C is arbitrary constant)

Choose the correct answer from the options given below

2022: 16 July Shift 2

Match List I with List II.

List I	List II
A. $\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$	I. order 3, degree 1
B. $\left(\frac{d^2y}{dx^2}\right)^2 = 0$	II. order 2, degree 2
C. $\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + y = 0$	III. order 2, degree 1
D. $\sin\left(\frac{dy}{dx}\right) + 5y = 0$	IV. order 1, degree is not defined

Choose the correct answer from the options given below:

2026: 19 May Shift 1

Match List-I with List-II

List-I (Differential equation)	List-II (The sum of degree and order of the differential equation)
(A) $x\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{3}}$	(I) 2
(B) $x\frac{d^2y}{dx^2} = \left(2-\frac{dy}{dx}\right)^{\frac{1}{7}}$	(II) 9
(C) $3+\frac{dy}{dx} = y+\frac{d^3y}{dx^3}$	(III) 4
(D) $\frac{dy}{dx} = x+3$	(IV) 5

Choose the correct answer from the options given below:

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