Question

\[\frac{dy}{dx}\] + y cot x = x² cot x + 2x

Answer 1

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

\[ \Rightarrow y \sin x = x^2 \int\cos xdx - \int\left[ \frac{d}{dx}\left( x^2 \right)\int\cos \text{ x } dx \right]dx + \int2x\sin \text{ x } dx + C\]
\[ \Rightarrow y \sin x = x^2 \sin x - \int2x\sin \text{ x }dx + \int2x\sin \text{ x }dx + C\]
\[ \Rightarrow y \sin x = x^2 \sin x + C\]
\[\text{ Hence, }y \sin x = x^2 \sin x + C\text{ is the required solution . }\]