The general solution of the differential equation (1 + e^x)dy + ye^x dx = 0, where y > 0, is Option 1: y = C(1 + e^x), C is an arbitary constant Option 2: C/e^x = y/(1 + e^x), C is an arbitary constant Option 3: C = y(1 + e^x), C is an arbitary constant Option 4: y = Ce^x/(1 + e^x), C is an arbitary constant

Answer: C = y(1 + e^x), C is an arbitary constant. Solution: Given: (1 + e^x)dy + ye^x dx = 0 where y > 0 This is a separable differential equation. (1 + e^x)dy = -ye^x dx dy/y = (-e^x dx)/(1 + e^x) Integrating both sides: dy/y = (-e^x)/(1 + e^x) dx y = -(1 + e^x) + C y + (1 + e^x) = C [y(1 + e^x)] = C y(1 + e^x) = C Therefore, the general solution is C = y(1 + e^x) where C is an arbitrary constant.

Calculus

Differential Equations

Medium

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The general solution of the differential equation $(1 + e^x)dy + ye^x dx = 0$ , where $y > 0$ , is

$y = C(1 + e^x)$ , C is an arbitary constant

$\frac{C}{e^x} = \frac{y}{1 + e^x}$ , C is an arbitary constant

$C = y(1 + e^x)$ , C is an arbitary constant

$y = \frac{Ce^x}{1 + e^x}$ , C is an arbitary constant

✅ Correct Option: 3

Given: $(1 + e^x)dy + ye^x dx = 0$ where $y > 0$

This is a separable differential equation.

$(1 + e^x)dy = -ye^x dx$

$\frac{dy}{y} = \frac{-e^x dx}{1 + e^x}$

Integrating both sides:

$\int \frac{dy}{y} = \int \frac{-e^x}{1 + e^x} dx$

$\ln y = -\ln(1 + e^x) + \ln C$

$\ln y + \ln(1 + e^x) = \ln C$

$\ln[y(1 + e^x)] = \ln C$

$y(1 + e^x) = C$

Therefore, the general solution is $C = y(1 + e^x)$ where $C$ is an arbitrary constant.

2025: 16 May Shift 1

The particular solution of the differential equation x(1 + y²)dx - y(1 + x²)dy = 0, y(0) = 1, is:

2026: 19 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $y + \frac{dy}{dx} = \frac{1}{4}\int y\,dx$	(I) Order 1, Degree 1
(B) $y = \frac{dy}{dx} + \frac{C}{dy/dx}$	(II) Order 2, Degree 2
(C) $(xy^2+x)dx + (y-x^2)dy = 0$	(III) Order 2, Degree 1
(D) $\left[\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{3/2}/\frac{d^2y}{dx^2}\right] = K$	(IV) Order 1, Degree 2

Choose the correct answer from the options given below:

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:

2026: 30 May Shift 2

The general solution of the differential equation $x(x^2-1)\frac{dy}{dx} = 1$ is:

[Where C is an arbitrary constant]

2025: 22 May Shift 2

Match List-I with List-II

List-I	List-II
Differential Equation	Order and Degree
(A) $\left(\frac{d^2y}{dx^2}\right)^2 = e^x\left(\frac{dy}{dx}\right)^4 + 1 = 0$	(I) order = 1 and degree = 2
(B) $\left(\frac{dy}{dx}\right)^2 + xy = 0$	(II) order = 2 and degree = 1
(C) $\left(1 + \frac{dy}{dx}\right)^{3/2} = 4\left(\frac{d^2y}{dx^2}\right)^2$	(III) order = 2 and degree = 2
(D) $\sqrt\frac{d^2y}{dx^2} + 1 = \frac{dy}{dx}$	(IV) order = 2 and degree = 4

Choose the correct answer from the options given below:

2026: 30 May Shift 1

The particular solution of the differential equation $\frac{dy}{dx}=x(2\log x+1)$ , given that $y=0$ when $x=2$ , is:

(consider $\log x=\log_e x$ )

The general solution of the differential equation (1+ex)dy+yexdx=0(1 + e^x)dy + ye^x dx = 0(1+ex)dy+yexdx=0, where y>0y > 0y>0, is

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CUET Mathematics Past Year Questions (Topic-Wise)

The general solution of the differential equation $(1 + e^x)dy + ye^x dx = 0$ , where $y > 0$ , is