CUET CUET Mathematics 2024 - The degree of the differential equation (1-(d y/d x)^2)^3/2=k d^2 yd x^2 is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Differential Equations

Easy

The degree of the differential equation $\left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}=k \frac{d^{2} y}{d x^{2}}$ is :

1

2

3

\frac{3}{2}

Correct Option: 2

The degree of the differential equation

\left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}=k \frac{d^{2} y}{d x^{2}}

2

First convert the equation to a polynomial form by eliminating all radicals (square both sides):

\left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{3}=k^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}

The highest power of the highest-order derivative

\left(\frac{d^{2} y}{d x^{2}}\right)

2

. Thus, the degree is

2

Note: Focus on the exponent of the highest-order derivative after removing radicals/fractions. Here, squaring removes the

\frac{3}{2}

exponent, leaving

\left(\frac{d^{2} y}{d x^{2}}\right)^{2}

, so degree

=2

Related questions:

CUET Mathematics 2024

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) $x^3$