Question

Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]

Answer 1

We have, 
\[\frac{dy}{dx} - 3y \cot x = \sin 2x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = - 3\cot x\text{ and }Q = \sin 2x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{- 3\int\cot x dx} \]
\[ = e^{- 3\log\left| \sin x \right|} = {cosec}^3 x\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = {\text{ cosec }}^3 x,\text{ we get }\]
\[ {\text{ cosec }}^3 x\left( \frac{dy}{dx} - 3y \cot x \right) = \sin 2x\left( {\text{ cosec }}^3 x \right)\]
\[ \Rightarrow {\text{ cosec }}^3 x\left( \frac{dy}{dx} - 3y \cot x \right) = 2\cot x\text{ cosec }x\]
Integrating both sides with respect to x, we get
\[y {\text{ cosec }}^3 x = 2\int\cot x\text{ cosec }x dx + C\]
\[ \Rightarrow y {\text{ cosec }}^3 x = - 2\text{ cosec }x + C\]
\[ \Rightarrow y = - 2 \sin^2 x + C \sin^3 x . . . . . \left( 2 \right)\]
Now,
\[y\left( \frac{\pi}{2} \right) = 2\]
\[ \therefore 2 = - 2 \sin^2 \frac{\pi}{2} + C \sin^3 \frac{\pi}{2}\]
\[ \Rightarrow C = 4\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y = - 2 \sin^2 x + 4 \sin^3 x\]
\[ \Rightarrow y = 4 \sin^3 x - 2 \sin^2 x\]
\[\text{ Hence, }y = 4 \sin^3 x - 2 \sin^2 x\text{ is the required solution.}\]