Question

$\frac{dy}{dx}+2y=x{e}^{4x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+2y=x{e}^{4x}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=2 \\ Q=x{e}^{4x} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int 2dx} \\ =e^{2x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}{e}^{2x},\mathrm{we}\mathrm{get} \\ {e}^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}\times x{e}^{4x} \\ \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=x{e}^{6x} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ {e}^{2x}y=\int \underset{\mathrm{II}}{e^{6x}}\underset{\mathrm{I}}{x}dx+C \\ \Rightarrow e^{2x}y=x\int e^{6x}dx-\int \left[\frac{d}{dx}\left(x\right)\int e^{6x}dx\right]dx+C \\ \Rightarrow e^{2x}y=\frac{x{e}^{6x}}{6}-\frac{e^{6x}}{36}+C \\ \Rightarrow y=\frac{x{e}^{4x}}{6}-\frac{e^{4x}}{36}+C{e}^{-2x} \\ \mathrm{Hence},y=\frac{x{e}^{4x}}{6}-\frac{e^{4x}}{36}+C{e}^{-2x}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$