(a)OR(b)

Solve the following differential equation : \[\left(y-\sin ^{2} x\right) d x+\tan x d y=0 \]

\[\begin{aligned} & \left(y-\sin ^{2} x\right) d x+\tan x d y=0 \\ & \frac{d y}{d x}+\frac{y}{\tan x}=\frac{\sin ^{2} x}{\tan x} \\ & \frac{d y}{d x}+(\cot x) y=\sin x \cos x \\ & \text { I.F. }=e^{\int \cot x d x}=e^{\log (\sin x)}=\sin x \end{aligned} \] Solution is given by \[y \sin x=\int \sin ^{2} x \cos x d x \] \(\displaystyle y \sin x=\int t^{2} d t \quad \because \sin x=t, \cos x d x=d t\)