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

The answer is 'C = y(1 + e^x), C is an arbitary constant'. View full solution:

Calculus > Differential Equations

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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

Related questions:

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

2026: 19th 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

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2026: 22nd 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 $y = (x-1)\log(x+1) - x + 5; y(0) = 5$

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2026: 30th 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: 30th 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 $\log x=\log_e x$ )