Question

$\frac{dy}{dx}+y=\cos x$

Answer 1

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