JEE Mathematics 2025 Question with Solution

Given:

$$x(x^2+e^x)\,dy+\big(e^x(x-2)y-x^3\big)\,dx=0$$

with the condition that the curve passes through $(1,0)$.

Find: $y(2)$.

Divide by $x(x^2+e^x)$ to write the equation in linear form:

$$\frac{dy}{dx}+\frac{e^x(x-2)}{x(x^2+e^x)}\,y=\frac{x^2}{x^2+e^x}.$$

So this is of the form $y' + P(x)y = Q(x)$ with

$$P(x)=\frac{e^x(x-2)}{x(x^2+e^x)},\qquad Q(x)=\frac{x^2}{x^2+e^x}.$$

Now find the integrating factor:

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \frac{e^x(x-2)}{x(x^2+e^x)}\,dx}.$$

Let

$$t=1+\frac{e^x}{x^2}.$$

Then

$$dt=\frac{e^x(x-2)}{x^3}\,dx.$$

Hence

$$\int \frac{e^x(x-2)}{x(x^2+e^x)}\,dx =\int \frac{\tfrac{e^x(x-2)}{x^3}}{\tfrac{x^2+e^x}{x^2}}\,dx =\int \frac{dt}{t} =\ln t.$$

Therefore,

$$\text{I.F.}=e^{\ln t}=1+\frac{e^x}{x^2}.$$

Multiply the differential equation by the integrating factor:

$$y\left(1+\frac{e^x}{x^2}\right)=\int Q(x)\cdot \text{I.F.}\,dx + C.$$

Now,

$$\int \frac{x^2}{x^2+e^x}\left(1+\frac{e^x}{x^2}\right)\,dx =\int 1\,dx =x+C.$$

So,

$$y\left(1+\frac{e^x}{x^2}\right)=x+C.$$

Use the point $(1,0)$:

$$0\cdot(1+e)=1+C.$$

Thus,

$$C=-1.$$

Therefore,

$$y=\frac{x-1}{1+\dfrac{e^x}{x^2}}.$$

Now substitute $x=2$:

$$y(2)=\frac{1}{1+\dfrac{e^2}{4}}=\frac{4}{4+e^2}.$$

Therefore, the correct option is D.

After converting the equation into linear form, the key observation is that the integrating factor is

$$1+\frac{e^x}{x^2}=\frac{x^2+e^x}{x^2}.$$

Then the right-hand side becomes

$$Q(x)\cdot \text{I.F.}=\frac{x^2}{x^2+e^x}\cdot \frac{x^2+e^x}{x^2}=1.$$

So the integral becomes immediately

$$\int 1\,dx=x+C.$$

This gives the compact form

$$y\left(1+\frac{e^x}{x^2}\right)=x+C,$$

from which the initial condition gives $C=-1$ and finally

$$y(2)=\frac{4}{4+e^2}.$$