Question

$\frac{dy}{dx}-y\tan x=e^x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-y\tan x=e^x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-\tan x \\ Q=e^x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int -\tan xdx} \\ =e^{-\log \left|\left(\sec x\right)\right|} \\ =e^{\log \left|\left(\cos x\right)\right|} \\ =\cos x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\cos x=\int e^x\cos xdx+C \\ \Rightarrow y\cos x=I+C.....\left(1\right) \\ \mathrm{Where}, \\ I=\int \underset{\mathrm{II}}{e^x}\underset{\mathrm{I}}{\cos x}dx.....\left(2\right) \\ \Rightarrow I=\cos x\int e^{x}dx-\int \left[\frac{d}{dx}\left(\cos x\right)\int e^{x}dx\right]dx \\ \Rightarrow I=\cos x{e}^x+\int \sin x{e}^{x}dx \\ \Rightarrow I=\cos x{e}^x+\int \underset{\mathrm{I}}{\sin x}\underset{\mathrm{II}}{e^x}dx \\ \Rightarrow I=\cos x{e}^x+\sin x\int e^{x}dx-\int \left[\frac{d}{dx}\left(\sin x\right)\int e^{x}dx\right]dx \\ \Rightarrow I=\cos x{e}^x+\sin x{e}^x-\int \cos x{e}^{x}dx \\ \Rightarrow I=\cos x{e}^x+\sin x{e}^x-I\left[\mathrm{From}\left(2\right)\right] \\ \Rightarrow 2I=\cos x{e}^x+\sin x{e}^x \\ \Rightarrow I=\frac{e^x}{2}\left(\cos x+\sin x\right) \\ \therefore y\cos x=\frac{e^x}{2}\left(\cos x+\sin x\right)+C\left[\mathrm{From}\left(1\right)\right] \end{gathered}$$