Let $y=Y(x)$ be the solution of the differential equation $\frac{d y}{d x}+y \tan x=2 x+x^2 \tan x$, $x…
Let $y=Y(x)$ be the solution of the differential equation $\frac{d y}{d x}+y \tan x=2 x+x^2 \tan x$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$, such that $Y(0)=1$, then
$\frac{d y}{d x}+y \tan x=2 x+x^2 \tan x, x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
On comparing with form $\frac{d y}{d x}+P y=Q$, where $P$ and $Q$ are the functions of $x$.
$
\therefore \quad P=\tan x \text { and } Q=2 x+x^2 \tan x
$
Hence, $I F=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$
$
y \cdot I F=\int Q \cdot I F+C
$
$\begin{aligned} & y \sec x=\int\left(2 x+x^2 \tan x\right) \sec x+C \\ & y \sec x=2 \int x \sec x d x+\int \frac{x^2 \sin x}{\cos ^2 x} d x+C \\ & y \sec x=\int 2 x \sec x d x+\int x^2 \tan x \cdot \sec x d x+C \\ & \Rightarrow y \sec x=x^2 \sec x+C \\ & \Rightarrow y=x^2+C \cos x...(i)\end{aligned}$
Now, when x = 0, y = 1 or y(0) = 1
From Eq. (i), we get $1=0+C \cos 0^{\circ}$
$
\begin{aligned}
& \Rightarrow 1=C \\
& \therefore y=x^2+\cos x \\
& \Rightarrow y^{\prime}=2 x-\sin x
\end{aligned}
$
$\begin{aligned} & \therefore y^{\prime}\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4}-\sin \frac{\pi}{4}=\frac{\pi}{2}-\frac{1}{\sqrt{2}} \\ & y^{\prime}\left(\frac{-\pi}{4}\right)=2\left(\frac{-\pi}{4}\right)-\sin \left(\frac{-\pi}{4}\right)=\frac{-\pi}{2}+\frac{1}{\sqrt{2}} \\ & \begin{aligned} \therefore y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(\frac{-\pi}{4}\right) & =\frac{\pi}{2}-\frac{1}{\sqrt{2}}-\left(\frac{-\pi}{2}+\frac{1}{\sqrt{2}}\right) \\ & =\frac{\pi}{2}+\frac{\pi}{2}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ \Rightarrow y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(\frac{-\pi}{4}\right) & =\pi-\sqrt{2}\end{aligned}\end{aligned}$