Question

\[\left( \sin x \right)\frac{dy}{dx} + y \cos x = 2 \sin^2 x \cos x\]

Answer 1

We have,
\[\left( \sin x \right)\frac{dy}{dx} + y \cos x = 2 \sin^2 x \cos x\]
\[ \Rightarrow \frac{dy}{dx} + y \cot x = 2\sin x \cos x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = \cot x\]
\[Q = 2\sin x \cos x\]
\[ \therefore I.F . = e^{\int{P dx}} \]
\[ = e^{ \int\cot \text{ x }dx } \]
\[ = e^{log\left| \sin x \right|} = \sin x\]
\[\text{Multiplying both sides of }\left( 1 \right)\text{ by }\sin x, \text{ we get }\]
\[\sin x\left( \frac{dy}{dx} + y \cot x \right) = \sin x \times 2\sin x\cos x\]
\[ \Rightarrow \sin x\frac{dy}{dx} + y \cos x = 2 \sin^2 x\cos x\]
Integrating both sides with respect to x, we get
\[y \sin x = 2\int \sin^2 x \cos \text{ x }dx + C . . . . . \left( 2 \right)\]
\[\text{ Putting }\sin x = t\]
\[ \Rightarrow \cos \text{ x } dx = dt\]
\[\text{ Therefore, }\left( 2 \right)\text{ becomes }\]
\[y \sin x = 2\int t^2 dt + C\]
\[ \Rightarrow y \sin x = \frac{2}{3} t^3 + C\]
\[ \Rightarrow y \sin x = \frac{2}{3} \sin^3 x + C\]
\[\text{ Hence, }y \sin x = \frac{2}{3} \sin^3 x + C\text{ is the required solution.} \]