Question

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

Answer 1

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