Calculus

Differential Equations

Easy

common

The general solution of the given differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^{x}+e^{-x}}$ (where C is an arbitrary constant)

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

$y = \frac{1}{3}e^{3x} + C$

$y = 9e^{3x} + C$

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

✅ Correct Option: 4

Write the denominator as $(e^{2x}+1)/e^{x}$ and factor the numerator as $3e^{2x}(1+e^{2x})$ . The bracket cancels, leaving $dy/dx = 3e^{3x}$ . Integrating gives $y = e^{3x} + C$ .

2023: 30 May Shift 3

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 :

2022: 10 Aug Shift 1

The integrating factor of differential equation $x\frac{dy}{dx} + 2y = x^2 \log x$ is

2022: 30 Aug Shift 1

The order and degree of the differential equation $\left[\left(\frac{d^2y}{dx^2}\right)^2 - 3\right]^{\frac{1}{3}} = 2\left(\frac{dy}{dx}\right)^{\frac{1}{4}}$ are

2025: 21 May Shift 1

The number of arbitrary constants in the general solution of a differential equation with degree 1 and order 3, is

2025: 21 May Shift 2

The solution of the differential equation $\frac{dy}{dx} = \frac{ax + c}{by + d}$ represents a circle when

2022: 17 Aug Shift 2

Consider the differential equation $\frac{dy}{dx} = \frac{y+1}{x+1}$ , and $y=0$ when $x=2$ . The value of $y$ at $x=3$ is :

The general solution of the given differential equation dydx=3e2x+3e4xex+e−x\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^{x}+e^{-x}}dxdy​=ex+e−x3e2x+3e4x​ (where C is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

The general solution of the given differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^{x}+e^{-x}}$ (where C is an arbitrary constant)