Let $y=y(x)$ be the solution of the differential equation

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

Then the area enclosed by the curve $f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y-x=4$ is ________.

<p>$$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$$</p> <p>I.F. of linear differential equation,</p> <p>$$\begin{aligned} & \text { I.F. }=e^{\int \frac{2 x}{\left(1+x^2\right)^2}} d x=e^{\left(\frac{-1}{1+x^2}\right)} \\ & \Rightarrow y\left(e^{\left(\frac{-1}{1+x^2}\right)}\right)=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\left(\frac{-1}{1+x^2}\right)} d x \\ & =\frac{x^2}{2}+c \\ & \Rightarrow y(0)=0 \Rightarrow 0\left(e^{-1}\right)=c \Rightarrow c=0 \\ & \Rightarrow y=\frac{e^{\frac{1}{1+x^2}} \cdot x^2}{2} \end{aligned}$$</p> <p>Area between curve $y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$ and $y-x=4$</p> <p>$$\begin{aligned} & \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow(x-4)(x+2)=0 \\ & \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\ & =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\ & =30-12=18 \end{aligned}$$</p>