Question

$x\frac{dy}{dx}-y=\left(x-1\right)e^x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}-y=\left(x-1\right)e^x \\ \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\left(\frac{x-1}{x}\right)e^x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=-\frac{1}{x} \\ Q=\left(\frac{x-1}{x}\right)e^x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int \frac{1}{x}dx} \\ =e^{-\log x} \\ =\frac{1}{x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\frac{1}{x},\mathrm{we}\mathrm{get} \\ \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}\left(\frac{x-1}{x}\right)e^x \\ \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=\left(\frac{x-1}{x^2}\right)e^x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{1}{x}y=\int \left(\frac{1}{x}-\frac{1}{x^2}\right)e^{x}dx+C \\ \Rightarrow \frac{1}{x}y=\frac{e^x}{x}+C \\ \Rightarrow y=e^x+Cx \\ \mathrm{Hence},y=e^x+Cx\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$