Question

Solve the following initial value problem:-

\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]

Answer 1

We have, 
\[\frac{dy}{dx} + 2y \tan x = \sin x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = 2\tan x\text{ and }Q = \sin x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{2\int\tan x dx} \]
\[ = e^{2\log\left| \sec x \right|} = \sec^2 x\]
\[\text{Multiplying both sides of }(1)\text{ by }I.F. = \sec^2 x, \text{ we get }\]
\[ \sec^2 x \left( \frac{dy}{dx} + 2y \tan x \right) = \sec^2 x \times \sin x\]
\[ \Rightarrow \sec^2 x \left( \frac{dy}{dx} + 2y \tan x \right) = \tan x \sec x\]
Integrating both sides with respect to x, we get
\[y \sec^2 x = \int\tan x \sec x dx + C\]
\[ \Rightarrow y \sec^2 x = \sec x + C . . . . . \left( 2 \right)\]
Now, 
\[y\left( \frac{\pi}{3} \right) = 0\]
\[ \therefore 0 \left( \sec\frac{\pi}{3} \right)^2 = \sec\frac{\pi}{3} + C\]
\[ \Rightarrow C = - 2\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y \sec^2 x = \sec x - 2\]
\[ \Rightarrow y = \cos x - 2 \cos^2 x\]
\[\text{ Hence, }y = \cos x - 2 \cos^2 x\text{ is the required solution.}\]