General solution of the differential equation (e^x + e^(-x))dy = (3e^2x + 3e^4x)dx is: [Where C is an arbitrary constant] Option 1: y = e^x + C Option 2: y = 3e^x + C Option 3: y = e^3x + C Option 4: y = 3e^3x + C

Answer: y = e^3x + C. Solution: RHS =3e^2x(1+e^2x) and LHS factor e^x+e^(-x)=e^(-x)(e^2x+1). So dy/dx=(3e^2x(1+e^2x))/(e^(-x)(1+e^2x))=3e^3x. Integrating, y=e^3x+C.

Calculus

Differential Equations

Medium

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General solution of the differential equation $(e^x + e^{-x})dy = (3e^{2x} + 3e^{4x})dx$ is:

[Where C is an arbitrary constant]

$y = e^x + C$

$y = 3e^x + C$

$y = e^{3x} + C$

$y = 3e^{3x} + C$

✅ Correct Option: 3

RHS $=3e^{2x}(1+e^{2x})$ and LHS factor $e^x+e^{-x}=e^{-x}(e^{2x}+1)$ .

So $\dfrac{dy}{dx}=\dfrac{3e^{2x}(1+e^{2x})}{e^{-x}(1+e^{2x})}=3e^{3x}$ .

Integrating, $y=e^{3x}+C$ .

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The general solution of the differential equation $\frac{xdy}{dx} + 4y = x^3, (x \neq 0)$ is:

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The general solution of the differential equation $\frac{dy}{dx} = e^{x-y} + x^2e^{-y}$ is equal to:

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Match List - I with List - II. Match the integrating factors :

List - I (Differential Equation)	List - II (Integrating factor)
(A) $\frac{dy}{dx} + 3y = e^{-2x}$	(I) $\frac{1}{x}$
(B) $x\frac{dy}{dx} + y = 3x^2$	(II) $e^{-x}$
(C) $x\frac{dy}{dx} - y = 3x^2$	(III) $x$
(D) $\frac{dy}{dx} - y = x$	(IV) $e^{3x}$

Choose the correct answer from the options given below :

General solution of the differential equation (ex+e−x)dy=(3e2x+3e4x)dx(e^x + e^{-x})dy = (3e^{2x} + 3e^{4x})dx(ex+e−x)dy=(3e2x+3e4x)dx is: [Where C is an arbitrary constant]

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CUET Mathematics Past Year Questions (Topic-Wise)

General solution of the differential equation $(e^x + e^{-x})dy = (3e^{2x} + 3e^{4x})dx$ is:

[Where C is an arbitrary constant]