Question

\[\frac{dy}{dx}\] = y tan x − 2 sin x

Answer 1

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