JEE Main 2025 — Mathematics Calculus

$$\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & x\left(x^2+e^x\right) \frac{d y}{d x}+e^x(x-2) y=x^3 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^2}{x^2+e^x} \\ & \text { I.F. }=e^{\int \frac{e^x(x-2)}{x\left(x^2+e^x\right)} d x}=e^{\int \frac{e^x\left(\frac{1}{x^2}-\frac{2}{x^2}\right) d x}{\left(1+\frac{e^x}{x^2}\right)} d x}\end{aligned}$$​x(x2+ex)dy+(ex(x−2)y−x3)dx=0x(x2+ex)dxdy​+ex(x−2)y=x3dxdy​+x(x2+ex)ex(x−2)​y=x2+exx2​ I.F. =e∫x(x2+ex)ex(x−2)​dx=e∫(1+x2ex​)ex(x21​−x22​)dx​dx​

Let $1+\frac{e^x}{x^2}=t \Rightarrow \frac{x^2 e^x-e^x 2 x}{x^4} d x=d t$1+x2ex​=t⇒x4x2ex−ex2x​dx=dt

$$\Rightarrow \text { I.F. } \mathrm{e}^{\ln \left(1+\frac{\mathrm{e}^2}{\mathrm{x}^2}\right)}=1+\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^2}$$⇒ I.F. eln(1+x2e2​)=1+x2ex​

Now $\displaystyle y\left(1+\frac{e^x}{x^2}\right)=\int \frac{x^2}{x^2+e^x} \cdot \frac{x^2+e^x}{x^2} d x+C$y(1+x2ex​)=∫x2+exx2​⋅x2x2+ex​dx+C

$$y\left(1+\frac{e^x}{x^2}\right)=x+C$$y(1+x2ex​)=x+C

Passing through $(1,0)$(1,0)

$\begin{aligned}
& \Rightarrow C=-1 \
& y=\frac{x-1}{1+\frac{e^x}{x^2}} \
& y(2)=\frac{1}{1+\frac{e^2}{4}}=\frac{4}{4+e^2}
\end{aligned}$