CUET Mathematics 2026 25th May Shift 1 PYQs

The general solution of the differential equation $x(x^2-1)\frac{dy}{dx} = 1$ is:

[Where C is an arbitrary constant]

Match the differential equation (List-I) with its order and degree (List-II)

LIST-I		LIST-II
A. $\dfrac{d^3y}{dx^3} + x^2\left(\dfrac{d^2y}{dx^2}\right) = 0$		I. order =3; degree is not defined
B. $\dfrac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 0$		II. order =3; degree=1
C. $1 - \left(\dfrac{dy}{dx}\right)^2 = \left(a\dfrac{d^2y}{dx^2}\right)^{1/3}$		III. order =2; degree=1
D. $y = x\left(\dfrac{dy}{dx}\right)^3 + \left(\dfrac{d^2y}{dx^2}\right)^4$		IV. order =2; degree=4

Choose the correct answer from the options given below:

Match the LIST-I with LIST-II

LIST-I (Differential Equation)		LIST-II (Degree of Differential Equation)
A. $\dfrac{d^3y}{dx^3} + \left(\dfrac{d^2y}{dx^2}\right)^2 - \dfrac{dy}{dx} + y = 0$		I. 2
B. $\dfrac{dy}{dx} + \left(\dfrac{d^2y}{dx^2}\right)^4 + 2\left(\dfrac{d^3y}{dx^3}\right)^3 = 0$		II. 4
C. $\left(\dfrac{d^3y}{dx^3}\right)^2 + \left(\dfrac{d^2y}{dx^2}\right)^4 - y = 0$		III. 1
D. $\left(\dfrac{d^2y}{dx^2}\right)^4 + \left(\dfrac{dy}{dx}\right)^3 + y = 0$		IV. 3

Choose the correct answer from the options given below: