Question

\[\frac{dy}{dx} - y \tan x = e^x\]

Answer 1

We have,

\[\frac{dy}{dx} - y \tan x = e^x \]

\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]

\[P = - \tan x \]

\[Q = e^x \]

Now,

\[I . F . = e^{\int - \tan x\ dx} \]

\[ = e^{- \log\left| \left( \sec x \right) \right|} \]

\[ = e^{\log\left| \left( \cos x \right) \right|} \]

\[ = \cos x\]

So, the solution is given by

\[y \cos x = \int e^x \cos x dx + C\]

\[ \Rightarrow y \cos\ x = I + C . . . . . . . . . . . \left( 1 \right)\]

Where,

\[ \Rightarrow I = \cos x\int e^x dx - \int\left[ \frac{d}{dx}\left( \cos x \right)\int e^x dx \right]dx\]

\[ \Rightarrow I = \cos x e^x + \int\sin x e^x dx\]

\[ \Rightarrow I = \cos x e^x + \sin x\int e^x dx - \int\left[ \frac{d}{dx}\left( \sin x \right)\int e^x dx \right]dx\]

\[ \Rightarrow I = \cos x e^x + \sin x e^x - \int\cos x e^x dx\]

\[ \Rightarrow I = \cos x e^x + \sin x e^x - I ............\left[\text{From (2)}\right]\]

\[ \Rightarrow 2I = \cos x e^x + \sin x e^x \]

\[ \Rightarrow I = \frac{e^x}{2}\left( \cos x + \sin x \right)\]

\[ \therefore y \cos x = \frac{e^x}{2}\left( \cos x + \sin x \right) + C .............\left[\text{From (1)} \right]\]