Question

Solve the following initial value problem:-

\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]

Answer 1

\[ \frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0 \]
\[\frac{dy}{dx} + y\cot x = 2\cos x . . . . \left( 1 \right) \]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = \cot x\text{ and }Q = 2\cos x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\cot x\ dx} \]
\[ = e^{\log{\sin x}} \]
\[ = \sin x\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = \sin x, \text{ we get }\]
\[\sin x\left( \frac{dy}{dx} + y\cot x \right) = 2\sin x\cos x\]
\[ \Rightarrow \sin x\frac{dy}{dx} + y\cos x = \sin2x\]
Integrating both sides with respect to x, we get
\[y\sin x = \int\sin 2x dx + C\]
\[ \Rightarrow y\sin x = - \frac{\cos2x}{2} + C . . . . . \left( 2 \right)\]
Now,
\[y\left( \frac{\pi}{2} \right) = 0 \]
\[ \therefore 0 \times \sin\left( \frac{\pi}{2} \right) = - \frac{cos\pi}{2} + C\]
\[ \Rightarrow C = - \frac{1}{2}\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y\sin x = - \frac{\cos2x}{2} - \frac{1}{2}\]
\[ \Rightarrow 2y\sin x = - \left( 1 + \cos2x \right)\]
\[ \Rightarrow 2y\sin x = - 2 \cos^2 x\]
\[ \Rightarrow y = - \cot x\cos x\]
\[\text{ Hence, }y = - \cot x\cos x\text{ is the required solution.}\]