Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \left(x^2+1\right)\frac{dy}{dx}-2xy=\left(x^4+2x^2+1\right)\cos x\\ & \frac{dy}{dx}-\left(\frac{2x}{x^2+1}\right)y=\frac{{\left(x^2+1\right)}^2\cos x}{{\mathrm{cx}}^2+1}=\left({\mathrm{x}}^2+1\right)\cos x\\ & \text{ (Linear D.E) }\\ & P=\frac{-2x}{x^2+1},Q=\left(x^2+1\right)\cos x\\ & \text{ I.F }=e^{\int Pdx}=e^{\int \frac{-2x}{x^2+1}dx}=\frac{1}{x^2+1}\\ & y\cdot \frac{1}{x^2+1}=\int \left(x^2+1\right)\cos x\cdot \frac{1}{x^2+1}dx\\ & \frac{y}{x^2+1}=\sin x+c\Rightarrow y\cos =1\Rightarrow c=1\\ & y=\left(x^2+1\right)(\sin x+1)\\ & {\int }_{-3}^{3}ydx={\int }_{-3}^3\left(x^2+1\right)(\sin x+1)\\ & dx={\int }_{-3}^3{x}^2\sin x+x^2\sin x+1dx\\ & \Rightarrow {\int }_{-3}^3{x}^2\sin xdx+{\int }_{-3}^3{x}^{2}dx+{\int }_{-3}^3\sin xdx+{\int }_{-3}^{3}1dx\\ & =0+18+0+6=24\end{array}$