Question

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

Answer 1

We have,

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

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

\[P = - \tan x \]

\[Q = e^x \sec x\]

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

\[y \cos\ x = \int\left( \cos x\ e^x \sec x \right) dx + C\]

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

\[ \therefore y \cos\ x = e^x + C\]