Question

`x cos x(dy)/(dx)+y(x sin x + cos x)=1`

Answer 1

We have,

\[x \cos x\frac{dy}{dx} + y \left( x \sin x + \cos x \right) = 1\]

\[ \Rightarrow \frac{dy}{dx} + \left( \tan x + \frac{1}{x} \right)y = \frac{1}{x \cos x}\]

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

\[P = \tan x + \frac{1}{x} \]

\[Q = \frac{1}{x \cos x}\]

Now,

\[I . F . = e^{\int\left( \tan x + \frac{1}{x} \right) dx} = e^{\log \left| x \sec x \right|} = x \sec x\]

So, the solution is given by

\[xy \sec x = \int \sec^2 x dx + C\]

\[ \Rightarrow xy \sec x = \tan x + C\]