Mathematics - Differential Equation Question with Solution | TestHub

$\frac{dy}{dx}+ay=x{e}^{\beta x}$ Integrating factor (I.F.) $={\mathrm{e}}^{\mathrm{adx}}={\mathrm{e}}^{\mathrm{ax}}$ So, the solution is $y\cdot e^{ax}=\int x{e}^{\beta x}\cdot e^{ax}dx$ $\Rightarrow y{e}^{ax}=\int x{e}^{(\alpha +\beta )x}dx$ If $\alpha +\beta \ne 0$ $\Rightarrow y{e}^{ax}=x\frac{e^{(\alpha +\beta )x}}{(\alpha +\beta )}-\frac{e^{(\alpha +\beta )x}}{(\alpha +\beta {)}^2}+C$ $\Rightarrow y=\frac{x{e}^{\beta x}}{(\alpha +\beta )}-\frac{e^{\beta x}}{(\alpha +\beta {)}^2}+C{e}^{-ax}$ $\Rightarrow y=\frac{e^{\beta x}}{(\alpha +\beta )}\left(x-\frac{1}{a+\beta }\right)+C{e}^{-\alpha x}$ Put $\alpha =\beta =1$ in (1) $y=\frac{e^x}{2}\left(x-\frac{1}{2}\right)+C{e}^{-x}$ $y(1)=1$ $1=\frac{e}{2}\times \frac{1}{2}+\frac{C}{e}\Rightarrow c=e-\frac{e^2}{4}$ So, $y=\frac{e^x}{2}\left(x-\frac{1}{2}\right)+\left(e-\frac{e^2}{4}\right)e^{-x}$ If $\alpha +\beta =0\&\alpha =1$ $\begin{array}{rl}& \frac{dy}{dx}+y=x{e}^{-x}\\ & \text{ I.F. }=e^x\end{array}$ Solution is $y{e}^x=\int xdx$ $\Rightarrow y{e}^x=\frac{x^2}{2}+C$ $y=\frac{x^2}{2}{e}^{-x}+C{e}^{-x}$ $y(1)=1$ $1=\frac{1}{2e}+\frac{C}{e}\Rightarrow C=e-\frac{1}{2}$ $y=\frac{x^2}{2}{e}^{-x}+\left(e-\frac{1}{2}\right)e^{-x}$