The particular solution of the differential equation dy/dx=x(2 x+1), given that y=0 when x=2, is: (consider x=_e x) Option 1: y=x x-32 Option 2: y=x^2 x-2 Option 3: y=x^2 x-42 Option 4: y=x^2 x+3

Answer: y=x^2 x-42. Solution: y=(2x x+x)dx=(x^2 x-x^2/2)+x^2/2=x^2 x+C. Using y=0 at x=2: 0=42+C, so C=-42 and y=x^2 x-42.

Calculus

Differential Equations

Medium

applied

The particular solution of the differential equation $\frac{dy}{dx}=x(2\log x+1)$ , given that $y=0$ when $x=2$ , is:

(consider $\log x=\log_e x$ )

$y=x\log x-3\log2$

$y=x^2\log x-\log2$

$y=x^2\log x-4\log2$

$y=x^2\log x+3$

✅ Correct Option: 3

$y=\int(2x\log x+x)dx=\left(x^2\log x-\frac{x^2}{2}\right)+\frac{x^2}{2}=x^2\log x+C$ . Using $y=0$ at $x=2$ : $0=4\log2+C$ , so $C=-4\log2$ and $y=x^2\log x-4\log2$ .

2023: 15 June Shift 2

The solution of the differential equation $\frac{dy}{dx} = \frac{6}{x^2}$ ; $y(1) = 3$ is :

2026: 25 May Shift 1

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

[Where C is an arbitrary constant]

2026: 18 May Shift 2

The general solution of the differential equation $\tan\left(\dfrac{(1+x^2)y'+2xy}{2x}\right)=1+x^2$ , is

(Consider $\log_e x=\log x$ )

(where C is an arbitrary constant)

2022: 25 May Shift 1

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

2023: 15 June Shift 2

Integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + xy = ay$ is :

2026: 29 May Shift 1

Match List-I with List-II

List-I (Differential Equation)	List-II (Degree)
(A) $\left(\dfrac{dy}{dx}\right)^3+\left(\dfrac{d^2y}{dx^2}\right)^2=0$	(I) Not defined
(B) $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^2+1=0$	(II) 2
(C) $\sqrt{1+\dfrac{d^2y}{dx^2}}=\dfrac{dy}{dx}+x$	(III) 1
(D) $\dfrac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0$	(IV) 3

Choose the correct answer from the options given below:

The particular solution of the differential equation dydx=x(2log⁡x+1)\frac{dy}{dx}=x(2\log x+1)dxdy​=x(2logx+1), given that y=0y=0y=0 when x=2x=2x=2, is: (consider log⁡x=log⁡ex\log x=\log_e xlogx=loge​x)

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CUET Mathematics Past Year Questions (Topic-Wise)

The particular solution of the differential equation $\frac{dy}{dx}=x(2\log x+1)$ , given that $y=0$ when $x=2$ , is:

(consider $\log x=\log_e x$ )