Question

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

Answer 1

$\frac{dy}{dx}+\frac{1}{\cos x}.y=\frac{\sin x}{\cos x}$
$\frac{1}{\cos x}.y=\tan x$
$\frac{dy}{dx}+y\sec x=\tan x$
Comparing with $\frac{dy}{dx}+py=q$
$P=\sec x,q=\tan x$
$I.f=e^{\int \sec xdx}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$
Therefore required solution is
$y.e^{\int Pdx}=\int e^{\int Pdx}Qdx$
$y(\sec x+x\tan x)=\int \tan x(\sec x+\tan x)dx+c$
$=\int (\sec x+\tan x)dx+\int {\tan }^{2}xdx+c=\sec x+\int ({\sec }^{2}x-1)dx+c$
$=\sec x+\tan x-x+c$