Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} + \frac{2x}{(1+x^2)^2} y = xe^{\frac{1}{(1+x^2)}}$$; $$y(0) = 0$$. Then the area enclosed by the curve $$f(x) = y(x)e^{-\frac{1}{(1+x^2)}}$$ and the line $$y - x = 4$$ is __________

The differential equation is $$\frac{dy}{dx} + \frac{2x}{(1+x^2)^2} y = xe^{\frac{1}{1+x^2}}$$ with $$y(0) = 0$$.

Find the integrating factor.

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

Let $$u = 1 + x^2$$, $$du = 2x \, dx$$:

$$ \int \frac{2x}{(1+x^2)^2} dx = \int \frac{du}{u^2} = -\frac{1}{u} = -\frac{1}{1+x^2} $$

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

Solve.

$$ y \cdot e^{-\frac{1}{1+x^2}} = \int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{-\frac{1}{1+x^2}} dx = \int x \, dx = \frac{x^2}{2} + C $$

Using $$y(0) = 0$$: $$0 = 0 + C$$, so $$C = 0$$.

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

Find $$f(x)$$.

$$ f(x) = y(x) \cdot e^{-\frac{1}{1+x^2}} = \frac{x^2}{2} $$

Find the area enclosed between $$f(x) = \frac{x^2}{2}$$ and $$y - x = 4$$ (i.e., $$y = x + 4$$).

Find intersection points: $$\frac{x^2}{2} = x + 4 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0$$.

$$x = -2$$ and $$x = 4$$.

$$ \text{Area} = \int_{-2}^{4} \left[(x + 4) - \frac{x^2}{2}\right] dx $$

$$ = \int_{-2}^{4} \left(x + 4 - \frac{x^2}{2}\right) dx = \left[\frac{x^2}{2} + 4x - \frac{x^3}{6}\right]_{-2}^{4} $$

At $$x = 4$$: $$8 + 16 - \frac{64}{6} = 24 - \frac{32}{3} = \frac{72-32}{3} = \frac{40}{3}$$.

At $$x = -2$$: $$2 - 8 + \frac{8}{6} = -6 + \frac{4}{3} = \frac{-18+4}{3} = -\frac{14}{3}$$.

$$ \text{Area} = \frac{40}{3} - \left(-\frac{14}{3}\right) = \frac{54}{3} = 18 $$

The answer is 18.