Official solution
\[{{dy} \over {dx}} + y(\tan x) = 2x + {x^2}\tan x\]
I.F. = \[{e^{\int {\tan x\,dx} }}\] = \[{e^{\ln \sec x}}\] = sec x
y. sec x = \[\int {(2x + {x^2}\tan x)\sec \,x\,dx} \]
\[ \Rightarrow y\sec x = {x^2}\sec x + \lambda \]
\[ \Rightarrow \] y(0) = 0 + \[\lambda \] = 1 \[ \Rightarrow \] \[\lambda \] = 1
\[ \Rightarrow \] y = x2 + cos x
\[ \Rightarrow \] y' = 2x -- sinx
\[ \Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}\]
\[ \Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}\]
\[ \therefore \] \[y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 \]