Question

\[\frac{dy}{dx}\] + y tan x = cos x

Answer 1

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