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

Answer: y = 2/(3-e^2x). Solution: Separating variables, e^(-y)dy=e^2xdx, so -e^(-y)=e^2x/2+C. At x=0,y=0: -1=12+C, giving C=-32. Then e^(-y)=(3-e^2x)/2, so y=, 2/(3-e^2x), .

Calculus

Differential Equations

Medium

core

Particular solution of the differential equation $dy = e^{2x+y}dx$ , $y(0) = 0$ is:

$y = \log\left|\frac{2}{3-e^{2x}}\right|$

$y = \log(3-e^{2x})$

$y = \frac{2}{3-e^{2x}}$

$y = 3-e^{2x}$

✅ Correct Option: 1

Separating variables, $e^{-y}dy=e^{2x}dx$ , so $-e^{-y}=\frac{e^{2x}}{2}+C$ .

At $x=0,y=0$ : $-1=\frac12+C$ , giving $C=-\frac32$ .

Then $e^{-y}=\frac{3-e^{2x}}{2}$ , so $y=\log\left|\frac{2}{3-e^{2x}}\right|$ .

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Match List-I with List-II

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

Choose the correct answer from the options given below:

Particular solution of the differential equation dy=e2x+ydxdy = e^{2x+y}dxdy=e2x+ydx, y(0)=0y(0) = 0y(0)=0 is:

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CUET Mathematics Past Year Questions (Topic-Wise)

Particular solution of the differential equation $dy = e^{2x+y}dx$ , $y(0) = 0$ is: