The general solution of the differential equation ydx + (x-y^3)dy = 0 is : Option 1: xy = y^4/4+cx : c is an arbitrary constant. Option 2: x = y^4/4+c : c is an arbitrary constant. Option 3: xy = y^4/4+c : c is an arbitrary constant. Option 4: xy = y^3/3+C : C is an arbitrary constant.

The answer is 'xy = y^4/4+c : c is an arbitrary constant.'. View full solution:

CUET Mathematics - The general solution of the differential equation ydx + (x-y^3)dy = 0 is : | PYQs + Solutions

CUET Mathematics 2026 12th May Shift 2

Calculus

Medium

core

$xy = \dfrac{y^4}{4}+cx$ : $c$ is an arbitrary constant.

$x = \dfrac{y^4}{4}+c$ : $c$ is an arbitrary constant.

$xy = \dfrac{y^4}{4}+c$ : $c$ is an arbitrary constant.

$xy = \dfrac{y^3}{3}+C$ : $C$ is an arbitrary constant.

✅ Correct Option: 3

Related questions:

16 May Shift 1

Choose the correct answer from the options given below:

List-I	List-II
(A) Integrating factor of $x \, dy - (y + 2x^2) \, dx = 0$	(I) $\frac{1}{x}$
(B) Integrating factor of $(2x^2 - 3y) \, dx = x \, dy$	(II) $x$
(C) Integrating factor of $(2y + 3x^2) \, dx + x \, dy = 0$	(III) $x^2$
(D) Integrating factor of $2x \, dy + (3x^3 + 2y) \, dx = 0$	(IV) $x^3$

30th May Shift 2

If $y = x\frac{dy}{dx} + \sqrt{a^2\left(\frac{dy}{dx}\right)^2 + b^2}$ , then the order and degree of the differential equation are

19th May Shift 1

Match List-I with List-II

List-I	List-II
(A) $y + \frac{dy}{dx} = \frac{1}{4}\int y\,dx$	(I) Order 1, Degree 1
(B) $y = \frac{dy}{dx} + \frac{C}{dy/dx}$	(II) Order 2, Degree 2
(C) $(xy^2+x)dx + (y-x^2)dy = 0$	(III) Order 2, Degree 1
(D) $\left[\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{3/2}/\frac{d^2y}{dx^2}\right] = K$	(IV) Order 1, Degree 2

Choose the correct answer from the options given below: