Given $\frac{dy}{dx}+y\tan x=2x+x^2\tan x$

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$ where $P=\tan x \& Q=2x+x^2\tan x$

Now, the integrating factor $I.F.=e^{\int Pdx}=e^{\int \tan x dx}$

$=e^{\mathrm{lnsec}x}=\sec x$

And, the general solution is $y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$

$y\cdot \sec x=\int \left(2x+x^2\tan x\right)\sec xdx+c$

$\Rightarrow ysecx=\int 2x\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{ }xdx+\int x^2\left(\sec x\cdot \tan x\right)dx+c$

Using integration by parts in the second integral, we get

$\Rightarrow ysecx=\int 2x\sec \mathrm{ }xdx+x^2\int \left(\sec x\cdot \tan x\right)dx-\int \left(\frac{d}{dx}\left(x^2\right)\int \left(\sec x\cdot \tan x\right)\right)dx+c$

$\Rightarrow ysecx=\int 2x\sec \mathrm{ }xdx+x^2\int secxdx-\int 2xsecxdx+c$

$\Rightarrow y\mathrm{s}\mathrm{e}\mathrm{c}x=x^2\mathrm{s}\mathrm{e}\mathrm{c}x+c$

$\Rightarrow y=x^2+c·\cos x$

Now, $y\left(0\right)=1$

$\Rightarrow 0+c=1\Rightarrow c=1$

So, $y=x^2+\mathrm{c}\mathrm{o}\mathrm{s}x$

Hence, $y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}+\frac{1}{\sqrt{2}}$ and $y\left(-\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}+\frac{1}{\sqrt{2}}$

Differentiating $y\left(x\right)$ with respect to $x$

$\Rightarrow y^{\text{'}}\left(x\right)=2x-\mathrm{s}\mathrm{i}\mathrm{n}x$

Hence, $y^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{\pi }{2}-\frac{1}{\sqrt{2}}$ and $y^{\text{'}}\left(-\frac{\pi }{4}\right)=-\frac{\pi }{2}+\frac{1}{\sqrt{2}}$

$\Rightarrow y^{\text{'}}\left(\frac{\pi }{4}\right)-y^{\text{'}}\left(-\frac{\pi }{4}\right)=\pi -\sqrt{2}.$