Question

$x\frac{dy}{dx}+y=x{e}^x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+y=x{e}^x \\ \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=e^x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=\frac{1}{x} \\ Q=e^x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\log x} \\ =x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}x,\mathrm{we}\mathrm{get} \\ x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x{e}^x \\ \Rightarrow x\frac{dy}{dx}+y=x{e}^x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ xy=\int x{e}^{x}dx+C \\ \Rightarrow xy=\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{e^x}dx+C \\ \Rightarrow xy=x\int e^{x}dx-\int \left(\frac{d}{dx}\left(x\right)\int e^{x}dx\right)dx+C \\ \Rightarrow xy=x{e}^x-e^x+C \\ \Rightarrow xy=\left(x-1\right)e^x+C \\ \Rightarrow y=\left(\frac{x-1}{x}\right)e^x+\frac{C}{x} \\ \mathrm{Hence},y=\left(\frac{x-1}{x}\right)e^x+\frac{C}{x}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$