Question

Solve the following initial value problem:-

\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]

Answer 1

\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
\[ \Rightarrow \frac{dy}{dx} = 2\cos x - ycot x \]
\[ \Rightarrow \frac{dy}{dx} + y\cot x = 2\cos x . . . . \left( 1 \right) \]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = \cot x\text{ and }Q = 2\cos x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\cot x\ dx} \]
\[ = e^{\log{sinx}} \]
\[ = \sin x\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = \sin x,\text{ we get }\]
\[\sin x\left( \frac{dy}{dx} + y\cot x \right) = 2\sin x\cos x\]
\[ \Rightarrow \sin x\frac{dy}{dx} + y\cos x = \sin2x\]
Integrating both sides with respect to x, we get
\[y\sin x = \int\sin 2x dx + C\]
\[ \Rightarrow y\sin x = - \frac{\cos2x}{2} + C \]
\[\text{ Hence, }y\sin x = - \frac{\cos2x}{2} + C\text{ is the required solution.}\]