Question

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

Answer 1

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