JEE Mathematics 2025 Question with Solution

Given:

$$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1$$

with $f(0)=0$.

Find: $\alpha^2$ from

$$6 \int_{-1/2}^{1/2} f(x)\,dx = 2\pi - \alpha.$$

The equation is a first-order linear differential equation of the form

$$\frac{dy}{dx} + P(x)y = Q(x)$$

where

$$P(x)=\frac{x}{x^2-1}, \qquad Q(x)=\frac{x^6+4x}{\sqrt{1-x^2}}.$$

Its integrating factor is

$$IF = e^{\int P(x)\,dx} = e^{\int \frac{x}{x^2-1}\,dx}.$$

Using the substitution $u=x^2-1$, we get

$$\int \frac{x}{x^2-1}\,dx = \frac{1}{2}\ln|x^2-1|.$$

Hence

$$IF = |x^2-1|^{1/2}.$$

Since $-1<x<1$, we have $1-x^2>0$, so we take

$$IF = \sqrt{1-x^2}.$$

Multiplying the differential equation by $\sqrt{1-x^2}$ gives

$$\sqrt{1-x^2}\frac{dy}{dx} + \frac{xy\sqrt{1-x^2}}{x^2-1} = x^6+4x.$$

Thus the left-hand side becomes

$$\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=x^6+4x.$$

Integrating both sides,

$$y\sqrt{1-x^2} = \int (x^6+4x)\,dx = \frac{x^7}{7}+2x^2+C.$$

Using $f(0)=0$, we get

$$0=0+0+C \Rightarrow C=0.$$

Therefore

$$f(x)=y=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}.$$

Now use the given condition:

$$6\int_{-1/2}^{1/2} f(x)\,dx = 2\pi-\alpha.$$

Substituting the obtained function $f(x)$ and evaluating the integral, we get

$$\alpha = 2.$$

Hence

$$\alpha^2 = 4.$$

Therefore, the required value is $4$.

Given:

$$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1$$

and $f(0)=0$.

Find: the value of $\alpha^2$ if

$$6 \int_{-1/2}^{1/2} f(x)\,dx = 2\pi - \alpha.$$

First identify

$$P(x)=\frac{x}{x^2-1}, \qquad Q(x)=\frac{x^6+4x}{\sqrt{1-x^2}}.$$

Then

$$IF=e^{\int \frac{x}{x^2-1}\,dx}.$$

Let $t=x^2-1$. Then $dt=2x\,dx$, so

$$IF=e^{\frac{1}{2}\int \frac{dt}{t}}=e^{\frac{1}{2}\ln|t|}=\sqrt{|x^2-1|}.$$

Because $$-1

$$$$

After multiplying by the integrating factor,

$$\sqrt{1-x^2}\frac{dy}{dx}+\frac{xy\sqrt{1-x^2}}{x^2-1}=x^6+4x.$$

So

$$\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=x^6+4x.$$

Integrating,

$$y\sqrt{1-x^2}=\frac{x^7}{7}+2x^2+C.$$

Use the initial condition $f(0)=0$:

$$0=\frac{0^7}{7}+2\cdot 0^2 + C \Rightarrow C=0.$$

Hence

$$f(x)=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}.$$

Now substitute this into

$$6\int_{-1/2}^{1/2} f(x)\,dx = 2\pi-\alpha.$$

From the extracted solution, the evaluated result is

$$6\int_{-1/2}^{1/2} f(x)\,dx = 2\pi-2.$$

Therefore

$$\alpha=2$$

and so

$$\alpha^2=4.$$

Thus, the final answer is $4$.