Question

Find the general solution of the differential equation $\frac{dy}{dx}-y=\cos x$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-y=\cos x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-1\mathrm{and}Q=\cos x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int dx} \\ =e^{-x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{-x},\mathrm{we}\mathrm{get} \\ {e}^{-x}\left(\frac{dy}{dx}-y\right)=e^{-x}\cos x \\ \Rightarrow e^{-x}\frac{dy}{dx}-e^{-x}y=e^{-x}\cos x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^{-x}=\int e^{-x}\cos xdx+C \\ \Rightarrow y{e}^{-x}=I+C.....\left(2\right) \\ \mathrm{Here}, \\ I=\int e^{-x}\cos xdx.....\left(3\right) \\ \Rightarrow I=e^{-x}\sin x-\int \left(-e^{-x}\sin x\right)dx \\ \Rightarrow I=e^{-x}\sin x+\int e^{-x}\sin xdx \\ \Rightarrow I=e^{-x}\sin x-e^{-x}\cos x-\int \left[\left(-e^{-x}\right)\times \left(-\cos x\right)\right]dx \\ \Rightarrow I=e^{-x}\sin x-e^{-x}\cos x-\int e^{-x}\cos xdx \\ \Rightarrow I=e^{-x}\sin x-e^{-x}\cos x-I\left[\mathrm{From}\left(3\right)\right] \\ \Rightarrow 2I=e^{-x}\left(\sin x-\cos x\right) \\ \Rightarrow I=\frac{e^{-x}}{2}\left(\sin x-\cos x\right).....\left(4\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(4\right)\mathrm{we}\mathrm{get} \\ \Rightarrow y{e}^{-x}=\frac{e^{-x}}{2}\left(\sin x-\cos x\right)+C \\ \Rightarrow y=\frac{1}{2}\left(\sin x-\cos x\right)+C{e}^x \\ \mathrm{Hence},y=\frac{1}{2}\left(\sin x-\cos x\right)+C{e}^x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$