Calculus

Differential Equations

Medium

core

The solution of $y' - y'' = 2x$ is:

A. $y = x^2 + 2x + 2$

B. $y = x^2 + 2x + 1$

C. $y = x + 2$

D. $y = x^2 - 2x + 1$

Choose the correct answer from the options given below:

A and B only

B only

C only

A and D only

✅ Correct Option: 1

Test each option in $y' - y'' = 2x$ . A: $y = x^2 + 2x + 2$ gives $y' - y'' = (2x+2) - 2 = 2x$ . True. B: $y = x^2 + 2x + 1$ gives $y' - y'' = 2x$ . True. C: $y = x + 2$ gives $1 - 0 = 1 \ne 2x$ . D: $y = x^2 - 2x + 1$ gives $(2x - 2) - 2 = 2x - 4 \ne 2x$ . So A and B only.

2026: 25 May Shift 1

Match the LIST-I with LIST-II

LIST-I Differential equation	LIST-II Integrating factor of the differential equation
A. $\dfrac{dy}{dx} + 2y = xe^{4x}$	I. $\dfrac{1}{x}$
B. $2x\dfrac{dy}{dx} + y = 6x^3$	II. $\dfrac{1}{y}$
C. $ydx - xdy + (\log_e x)dx = 0$	III. $e^{2x}$
D. $ydx - (x+2y^2)dy = 0$	IV. $\sqrt{x}$

Choose the correct answer from the options given below:

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:

2026: 21 May Shift 1

General solution of the Differential equation $\frac{dy}{dx} + ay = e^{mx}$ is

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
Differential Equation	General solution
---	---
(A) $\dfrac{dy}{dx} = \dfrac{y}{x}; x \neq 0$	(I) $y = cx; c \text{ is an arbitrary constant}$
(B) $x dx - y dy = 0; y \neq 0, x \neq 0$	(II) $x^2 - y^2 = c; c \text{ is an arbitrary constant}$
(C) $\dfrac{(x^2 - 1)}{y^2 + 1}\dfrac{dx}{dy} = 1$	(III) $2x + 3y = c; c \text{ is an arbitrary constant}$
(D) $2 dx + 3 dy = 0$	(IV) $(x^3-y^3) = c + 3(x+y); c \text{ is an arbitrary constant}$

Choose the correct answer from the options given below:

2026: 21 May Shift 2

consider the differential equation $\dfrac{dx}{dy} + Px = Q$ , where $P$ and $Q$ are function of $y$ . Then

A. Its integrating factor (I.F.) is $e^{\int Pdy}$

B. Its integrating factor (I.F.) is $e^{\int Pdx}$

C. Its solution is given by $x(I.F.) = \int Q(I.F.)dy + C$

D. Its solution is given by $y(I.F.) = \int Q(I.F.)dx + C$

Choose the correct answer from the options given below:

2025: 16 May Shift 1

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right)= 3x + 4y$ satisfying $y = 0$ when $x = 0$ is:

The solution of y′−y′′=2xy' - y'' = 2xy′−y′′=2x is: A. y=x2+2x+2y = x^2 + 2x + 2y=x2+2x+2 B. y=x2+2x+1y = x^2 + 2x + 1y=x2+2x+1 C. y=x+2y = x + 2y=x+2 D. y=x2−2x+1y = x^2 - 2x + 1y=x2−2x+1 Choose the correct answer from the options given below:

Solution

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Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The solution of $y' - y'' = 2x$ is:

A. $y = x^2 + 2x + 2$

B. $y = x^2 + 2x + 1$

C. $y = x + 2$

D. $y = x^2 - 2x + 1$

Choose the correct answer from the options given below: