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Question
`x cos x(dy)/(dx)+y(x sin x + cos x)=1`
Solution
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
\[xy \sec x = \int \sec^2 x dx + C\]
\[ \Rightarrow xy \sec x = \tan x + C\]
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