Question

\[\frac{dy}{dx} + y \tan x = x^2 \cos^2 x\]

Answer 1

We have,
\[\frac{dy}{dx} + y \tan x = x^2 \cos^2 x\]

\[\text{Comparing with }\frac{dy}{dx} + Py = Q, \text{ we get }\]
\[P = \tan x \]
\[Q = x^2 \cos^2 x\]
Now, 
\[I . F . = e^{\int\tan x\ dx} = e^{log \left| \sec x \right|} = \sec x\]
Therefore, solution is given by
\[y \times I . F . = \int x^2 \cos^2 x \times I . F . dx + C\]
\[ \Rightarrow y \sec x = \int x^2 \cos x dx + C\]
\[ \Rightarrow y \sec x = I + C\]
Where, 

\[ \Rightarrow I = x^2 \int\cos x dx - \int\left[ \frac{d}{dx}\left( x^2 \right)\int\cos x dx \right]dx\]
\[ \Rightarrow I = x^2 \sin x - 2\int x \sin x dx\]
\[ \Rightarrow I = x^2 \sin x - 2\int x_I \sin x_{II} dx\]
\[ \Rightarrow I = x^2 \sin x - 2x\int\sin x dx + 2\int\left[ \frac{d}{dx}\left( x \right)\int\sin x dx \right]dx\]
\[ \Rightarrow I = x^2 \sin x + 2x \cos x - 2\int\cos x dx\]
\[ \Rightarrow I = x^2 \sin x + 2x \cos x - 2\sin x\]
\[ \Rightarrow I = x^2 \sin x + 2x \cos x - 2\sin x\]
\[ \therefore y \sec x = x^2 \sin x + 2x \cos x - 2\sin x + C\]
\[ \Rightarrow y \sec x = x^2 \sin x + 2x \cos x - 2\sin x + C\]