The order and degree of differential equation y = px + sqrt(a^2p^2 + b^2), p = dy/dx, a and b are constants are: Option 1: order = 1, degree = 1 Option 2: order = 2, degree = 1/2 Option 3: order = 1, degree not defined Option 4: order = 1, degree = 2

The answer is 'order = 1, degree = 2'. View full solution:

CUET Mathematics - The order and degree of differential equation y = px + sqrt(a^2p^2 + b^2), p = dy/dx, a and b are constants are: | PYQs + Solutions

CUET Mathematics 2026 15th May Shift 1

Calculus

Medium

common

order = 1, degree = 1

order = 2, degree = 1/2

order = 1, degree not defined

order = 1, degree = 2

✅ Correct Option: 4

Related questions:

26th May Shift 2

Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Order and Degree
A.	$\left(\dfrac{dy}{dx}\right)^2 + 3y\left(\dfrac{d^2y}{dx^2}\right) = 0$	I.	$1,1$
B.	$\sqrt{1-\left(\dfrac{dy}{dx}\right)^2} = a\left(\dfrac{d^2y}{dx^2}\right)^{1/3}$	II.	$1,2$
C.	$y = \dfrac{dy}{dx} + \dfrac{5}{(dy/dx)}$	III.	$2,1$
D.	$\sqrt{1-y^2}\,dx + \sqrt{1-x^2}\,dy = 0$	IV.	$2,2$

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21st May Shift 2

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

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23 Aug Shift 1

Match List - I with List - II

List - I (Differential Equation)	List - II (Degree)
A. $\left[1 + (y')^2\right]^2 = y''$	I. $2$
B. $\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3$	II. $4$
C. $(y''')^2 + y'' + 3y' + 5y = e^x$	III. $1$
D. $\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2$	IV. $3$

Choose the correct answer from the option given below: