Calculus

Differential Equations

Medium

common

Let $y = f(x)$ be the solution curve of the differential equation $\dfrac{dy}{dx} + 2y^2 = 0$ if $f(1) = 1$ , then $f(2)$ is equal to

$-\dfrac{1}{4}$

$-\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{4}$

✅ Correct Option: 3

Separating, $dy/y^2 = -2\,dx$ , so $-1/y = -2x + c$ , i.e. $1/y = 2x + c_1$ .

Using $f(1)=1$ gives $1 = 2 + c_1$ , so $c_1 = -1$ and $1/y = 2x-1$ .

At $x=2$ , $f(2) = 1/3$ .

2026: 14 May Shift 2

The solution of the differential equation $\frac{dy}{dx} = x^2 + x + \frac{1}{x}$ is

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: 6 Aug Shift 2

The curve passing through the point $(-1, 1)$ , given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$ also passes through the point $\left( k, -\frac{1}{2} \right)$ , then

2026: 26 May Shift 2

The order and degree of the differential equation $\dfrac{d^3y}{dx^3} + \left(\dfrac{d^2y}{dx^2}\right)^2 + e^{\frac{dy}{dx}} = 2$ is

2025: 3 June Shift 2

Match List-I with List-II

List-I	List-II
(A) Degree of this differential equation $\frac{d^4y}{dx^4} + 2\log_e\left(\frac{d^3y}{dx^3}\right) = 0$	(I) 1
(B) Order of this differential equation $e^{\left(\frac{dy}{dx}\right)^3} + 3y\left(\frac{d^2y}{dx^2}\right)^3 = 0$	(II) 4
(C) Degree of $\frac{d^4y}{dx^4} + \left(\frac{dy}{dx}\right)^2 = 0$	(III) not defined
(D) Order of the differential equation $2\frac{d^4y}{dx^4} + \left(\frac{d^2y}{dx^2}\right)^5 = 0$	(IV) 2

Choose the correct answer from the options given below:

Let y=f(x)y = f(x)y=f(x) be the solution curve of the differential equation dydx+2y2=0\dfrac{dy}{dx} + 2y^2 = 0dxdy​+2y2=0 if f(1)=1f(1) = 1f(1)=1, then f(2)f(2)f(2) is equal to

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $y = f(x)$ be the solution curve of the differential equation $\dfrac{dy}{dx} + 2y^2 = 0$ if $f(1) = 1$ , then $f(2)$ is equal to