The general solution of the differential equation dy/dx = (3e^2x+3e^4x)/(e^x+e^(-x)) is Option 1: y = e^3x + C, where C is arbitrary constant Option 2: y = e^3x/3 + C, where C is arbitrary constant Option 3: y = 3e^ex + C, where C is arbitrary constant Option 4: y = e^2x/2 + C, where C is arbitrary constant

Answer: y = e^3x + C, where C is arbitrary constant. Solution: Numerator =3e^2x(1+e^2x); denominator =e^(-x)(1+e^2x). Cancelling (1+e^2x) gives dy/dx=3e^3x. Integrating, y=e^3x+C.

Calculus

Differential Equations

Medium

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The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}$ is

$y = e^{3x} + C$ , where C is arbitrary constant

$y = \frac{e^{3x}}{3} + C$ , where C is arbitrary constant

$y = 3e^{ex} + C$ , where C is arbitrary constant

$y = \frac{e^{2x}}{2} + C$ , where C is arbitrary constant

✅ Correct Option: 1

Numerator $=3e^{2x}(1+e^{2x})$ ; denominator $=e^{-x}(1+e^{2x})$ .

Cancelling $(1+e^{2x})$ gives $\frac{dy}{dx}=3e^{3x}$ .

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

2025: 19 May Shift 1

Match List-I with List-II

List-I	List-II
Differential Equation	Order and degree of differential equation
(A) $\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 = e^{\frac{dy}{dx}} + 1$	(I) Order = 1, Degree = 2
(B) $\left(\frac{d^2y}{dx^2}\right)^2 + 4\left(\frac{dy}{dx}\right)^3 = e^y - 1$	(II) Order = 2, Degree = 1
(C) $3\left(\frac{dy}{dx}\right) + 4y + e^y = \frac{dx}{dy}$	(III) Order = 2, Degree = 2
(D) $\frac{d^2y}{dx^2} + 3\left(\frac{dy}{dx}\right) = \left(e^y + \frac{dy}{dx}\right)^2$	(IV) Order = 2, Degree = Not defined

Choose the correct answer from the options given below:

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: 26 May Shift 2

The solution of the differential equation $xdy - ydx = 0$ represents

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
Differential Equation	General solution
---	---
(A) $\dfrac{dy}{dx} = \dfrac{y}{x}; x \neq 0$	(I) $y = cx; c \text{ is an arbitrary constant}$
(B) $x dx - y dy = 0; y \neq 0, x \neq 0$	(II) $x^2 - y^2 = c; c \text{ is an arbitrary constant}$
(C) $\dfrac{(x^2 - 1)}{y^2 + 1}\dfrac{dx}{dy} = 1$	(III) $2x + 3y = c; c \text{ is an arbitrary constant}$
(D) $2 dx + 3 dy = 0$	(IV) $(x^3-y^3) = c + 3(x+y); c \text{ is an arbitrary constant}$

Choose the correct answer from the options given below:

2022: 25 May Shift 1

The solution of the differentiable equation $2x\frac{dy}{dx} + y = 14x^3, x > 0$ , is

2025: 26 May Shift 2

In the following differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 2x^2 \log\left(\frac{d^2y}{dx^2}\right)$ order and degree is:

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

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CUET Mathematics Past Year Questions (Topic-Wise)

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}$ is