Question

Solve the differential equation:
$\frac{dy}{dx}+y\tan x={\cos }^{3}x$

Answer 1

$\frac{dy}{dx}+y\tan x={\cos }^{3}x$

Here $p=\tan x,q={\cos }^{3}x$

Integrating factor $=e^{\int pdx}=e^{\int \tan xdx}=e^{\ln \left(\sec x\right)}=\sec x$

Therefore,

Solution of the equation is-

$y(I.F.)=\int q\times I.F.dx$

$\Rightarrow y\cdot \sec x=\int {\cos }^{3}x\sec xdx$

$\Rightarrow y\cdot \sec x=\int {\cos }^{2}xdx$

$\Rightarrow y\cdot \sec x$ $=\int \left(\frac{1+\cos 2x}{2}\right)dx(\because \cos 2x=2{\cos }^{2}x-1)$

$\Rightarrow y=\frac{x\cdot \cos x}{2}+\frac{1}{4}\sin 2x\cdot \cos x+C\cos x$