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

First, let's identify that the highest-order derivative is

\frac{d^2y}{dx^2}

(second-order).

Next, we need to eliminate the radical by squaring both sides:

\newline

\begin{aligned} & {\left[\left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}\right]^{2}=\left[k \frac{d^{2} y}{d x^{2}}\right]^{2}} \\ & \left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{3}=k^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\end{aligned}

Now the equation is in polynomial form. The highest-order derivative

\frac{d^2y}{dx^2}

is raised to the power of 2.

Therefore, the degree of this differential equation is 2.

Note: The degree is determined by the highest power of the highest-order derivative after converting to polynomial form. Don't confuse this with the order of the equation (which is also 2 in this case).

Related questions:

CUET Mathematics 2024

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) $x^3$