Let y=y(x) be the solution of the differential equation (dy)/(dx)+2x(1+x^2)^2y=xe^1(1+x^2)…

Question

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

$\frac{dy}{dx}+\frac{2x}{{(1+{x}^{2})}^{2}}y=x{e}^{\frac{1}{(1+{x}^{2})}};y(0)=0.$

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

Answer: 18

Accepted Answer

$\frac{dy}{dx}+\frac{2x}{{(1+{x}^{2})}^{2}}y=x{e}^{\frac{1}{1+{x}^{2}}};y(0)=0$

I.F. of linear differential equation,

$\begin{gathered} ={e}^{\int \frac{2x}{{(1+{x}^{2})}^{2}}}dx={e}^{(\frac{-1}{1+{x}^{2}})} \ \Rightarrow y({e}^{(\frac{-1}{1+{x}^{2}})})=\int x\cdot {e}^{\frac{1}{1+{x}^{2}}}\cdot {e}^{(\frac{-1}{1+{x}^{2}})}dx \ =\frac{{x}^{2}}{2}+c \ \Rightarrow y(0)=0\Rightarrow 0({e}^{-1})=c\Rightarrow c=0 \ \Rightarrow y=\frac{{e}^{\frac{1}{1+{x}^{2}}}\cdot {x}^{2}}{2} \end{gathered}$

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

$\begin{gathered} \Rightarrow 2(x+4)={x}^{2}\Rightarrow {x}^{2}-2x-8=0 \ \Rightarrow (x-4)(x+2)=0 \ {\int }_{4}^{-2}[(x+4)-\frac{{x}^{2}}{2}]dx=\frac{{x}^{2}}{2}+4x-{\frac{{x}^{3}}{6}|}_{4}^{-2} \ =(8+16-\frac{64}{6})-(2-8+\frac{8}{6}) \ =30-12=18 \end{gathered}$

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