QJEEM24Mabfb8: JEE Main 2024 Mathematics - Class 12 JEEM24Mabfb8 Solution: Let be the solution of the differential equation Then the area…

$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$

I.F. of linear differential equation,

$\begin{aligned}& \text { I.F. }=e^{\int \frac{2 x}{\left(1+x^2\right)^2}} d x=e^{\left(\frac{-1}{1+x^2}\right)} \\& \Rightarrow y\left(e^{\left(\frac{-1}{1+x^2}\right)}\right)=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\left(\frac{-1}{1+x^2}\right)} d x \\& =\frac{x^2}{2}+c \\& \Rightarrow y(0)=0 \Rightarrow 0\left(e^{-1}\right)=c \Rightarrow c=0 \\& \Rightarrow y=\frac{e^{\frac{1}{1+x^2}} \cdot x^2}{2}\end{aligned}$

Area between curve $y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$ and $y-x=4$

$\begin{aligned}& \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\& \Rightarrow(x-4)(x+2)=0 \\& \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\& =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\& =30-12=18\end{aligned}$