Question

Solve the following differential equation:-

\[\frac{dy}{dx} + \left( \sec x \right) y = \tan x\]

Answer 1

We have,

\[\frac{dy}{dx} + \left( \sec x \right)y = \tan x\]

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

\[P = \sec x\]

\[Q = \tan x\]

Now,

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

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

\[ = \sec x + \tan x\]

So, the solution is given by

\[y \times I . F = \int Q \times I . F . dx + C\]

\[ \Rightarrow y\left( \sec x + \tan x \right) = \int\left( \sec x + \tan x \right)\tan x + C\]

\[ \Rightarrow y\left( \sec x + \tan x \right) = \int\sec x \times \tan x dx + \int \tan^2 x dx + C\]

\[ \Rightarrow y\left( \sec x + \tan x \right) = \int\sec x \times \tan x dx + \int\left( \sec^2 x - 1 \right) dx + C\]

\[ \Rightarrow y\left( \sec x + \tan x \right) = \sec x + \tan x - x + C\]