Particular solution of the differential equation (dy/dx) = x + y, given that when x = 0, y = 0 is: Option 1: e^x + e^(-y) = 2 Option 2: e^(-x) + e^y = 2 Option 3: e^x + e^y = 2 Option 4: e^(-x) + e^(-y) = 2

Answer: e^x + e^(-y) = 2. Solution: dy/dx = e^(x+y) = e^x * e^y. Separating: e^(-y) dy = e^x dx. Integrating: -e^(-y) = e^x + C. At (0,0): -1 = 1 + C, so C = -2. Thus e^x + e^(-y) = 2.

Calculus

Differential Equations

Medium

core

Particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x + y$ , given that when $x = 0, y = 0$ is:

$e^x + e^{-y} = 2$

$e^{-x} + e^y = 2$

$e^x + e^y = 2$

$e^{-x} + e^{-y} = 2$

✅ Correct Option: 1

$\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$ . Separating: $e^{-y}\,dy = e^x\,dx$ . Integrating: $-e^{-y} = e^x + C$ . At $(0,0)$ : $-1 = 1 + C$ , so $C = -2$ . Thus $e^x + e^{-y} = 2$ .

2025: 26 May Shift 2

Match List-I with List-II

List-I	List-II
Differential Equations	Order and degree
(A) $ydx + x\log(y/x)dy - 2xdy = 0$	(I) Order : 2, degree:1
(B) $\left(\frac{d^3y}{dx^3}\right)^2 + 3\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^2$	(II) Order :1, degree:1
(C) $\frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y$	(III) Order : 3, degree:2
(D) $\left(\frac{ds}{dt}\right)^4 + 2s\frac{d^2s}{dt^2} = 0$	(IV) Order : 1, degree: Not defined

Choose the correct answer from the options given below:

2026: 22 May Shift 1

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2022: 30 Aug Shift 1

Solution of the differential equation $(x + xy)dy - y(1 - x^2)dx = 0$ is

2025: 27 May Shift 1

Which one of the following equations is a homogeneous differential equation?

2025: 13 May Shift 1

Match List-I with List-II

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

Choose the correct answer from the options given below:

2025: 14 May Shift 2

General solution of the differential equation $\frac{dy}{dx} = e^{\frac{x^2}{2}} + xy$ is

Particular solution of the differential equation log⁡(dydx)=x+y\log\left(\frac{dy}{dx}\right) = x + ylog(dxdy​)=x+y, given that when x=0,y=0x = 0, y = 0x=0,y=0 is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x + y$ , given that when $x = 0, y = 0$ is: