Question

$\frac{dy}{dx}+y\tan x=x^2{\cos }^{2}x$

Answer 1

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