Question

Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]

Answer 1

We have,
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = \cot x\text{ and }Q = 4x \text{ cosec }x \]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\cot x dx} \]
\[ = e^{log\left| \sin x \right|} = \sin x\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = \sin x,\text{ we get }\]
\[\sin x\left( \frac{dy}{dx} + y \cot x \right) = \sin x\left( 4x\text{ cosec }x \right)\]
\[ \Rightarrow \sin x\left( \frac{dy}{dx} + y \cot x \right) = 4x\]
Integrating both sides with respect to x, we get
\[y \sin x = 4\int x dx + C\]
\[ \Rightarrow y \sin x = 2 x^2 + C . . . . . \left( 2 \right)\]
Now, 
\[y\left( \frac{\pi}{2} \right) = 0\]
\[ \therefore 0 \times \sin\left( \frac{\pi}{2} \right) = 2 \left( \frac{\pi}{2} \right)^2 + C\]
\[ \Rightarrow C = - \frac{\pi^2}{2}\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y \sin x = 2 x^2 - \frac{\pi^2}{2}\]
\[\text{ Hence, }y \sin x = 2 x^2 - \frac{\pi^2}{2}\text{ is the required solution .}\]