Question

$\frac{dy}{dx}+2y=\sin x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+2y=\sin x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=2 \\ \mathrm{and} \\ Q=\sin x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int 2dx} \\ =e^{2x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{2x},\mathrm{we}\mathrm{get} \\ {e}^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}\sin x \\ \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}\sin x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^{2x}=\int e^{2x}\sin xdx+C \\ \Rightarrow y{e}^{2x}=\frac{1}{5}\int \left[2e^{2x}\left(2\sin x-\cos x\right)+e^{2x}\left(2\cos x+\sin x\right)\right]dx+C \\ \mathrm{Putting}{e}^{2x}\left(2\sin x-\cos x\right)=t \\ \Rightarrow \left[2e^{2x}\left(2\sin x-\cos x\right)+e^{2x}\left(2\cos x+\sin x\right)\right]dx=dt \\ \therefore y{e}^{2x}=\frac{1}{5}\int dt+C \\ \Rightarrow y{e}^{2x}=\frac{t}{5}+C \\ y{e}^{2x}=\frac{e^{2x}}{5}\left(2\sin x-\cos x\right)+C \\ \Rightarrow y=\frac{1}{5}\left(2\sin x-\cos x\right)+C{e}^{-2x} \\ \mathrm{Hence},y=\frac{1}{5}\left(2\sin x-\cos x\right)+C{e}^{-2x}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$