The general solution of the differential equation x^2dy/dx + y^2 = xy is: (take _e x = x) Option 1: x/y = x + C, C is an arbitrary constant Option 2: x/y = 2 x + C, C is an arbitrary constant Option 3: y = x + C, C is an arbitrary constant Option 4: x = y + C, C is an arbitrary constant

The answer is 'x/y = x + C, C is an arbitrary constant'. View full solution:

CUET Mathematics - The general solution of the differential equation x^2dy/dx + y^2 = xy is: (take _e x = x) | PYQs + Solutions

CUET Mathematics 2026 31st May Shift 2

Calculus

Medium

core

$\dfrac{x}{y} = \log x + C$ , C is an arbitrary constant

$\dfrac{x}{y} = 2\log x + C$ , C is an arbitrary constant

$y = \log x + C$ , C is an arbitrary constant

$x = \log y + C$ , C is an arbitrary constant

✅ Correct Option: 1

Related questions:

15th May Shift 1

If $p$ and $q$ are degree and order of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^3 - 2\left(\dfrac{dy}{dx}\right)^4 = 6$ respectively. Then the value of $2p+3q$ is

22nd May Shift 1

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

30th May Shift 2

Match List-I with List-II

List-I	List-II
Curve $y=f(x)$ (where C is an arbitrary constant)	The curve $y=f(x)$ satisfies the differential equation
(A) $y = \log x + C$	(I) $x\frac{dy}{dx} - y = 0$
(B) $y = e^x + C$	(II) $x\frac{dy}{dx} - 1 = 0$
(C) $y = cx$	(III) $x^2\frac{dy}{dx} + 1 = 0$
(D) $y = \frac{1}{x} + C$	(IV) $\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

Choose the correct answer from the options given below:

CUET Mathematics 2026 31st May Shift 2 PYQs

Keyboard Shortcuts