If \(y(x)\) satisfies the differential equation \(y^{\prime}-y \tan x\) \(=2 x \sec x\) and \(y(0)=0\), then

\(\frac{d y}{d x}-y \tan x=2 x \sec x\)

I.F. \(=e^{-\int \tan x d x}=\cos x\)

\(\therefore \quad y \cdot \cos x=\int 2 x d x=x^{2}+c\)

Now, \(y(0)=0 \Rightarrow c=0, \therefore y=x^{2} \sec x\)

\(\Rightarrow y^{\prime}=2 x \sec x+x^{2} \sec x \tan x\)

Now, \(y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16} \times \sqrt{2}=\frac{\pi^{2}}{8 \sqrt{2}}\)

\(y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9} \times 2=\frac{2 \pi^{2}}{9}\)

\(y^{\prime}\left(\frac{\pi}{4}\right)=\frac{2 \pi}{4} \times \sqrt{2}+\frac{\pi^{2}}{8 \sqrt{2}} \times 1=\frac{\pi^{2}}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}}\)

\(y^{\prime}\left(\frac{\pi}{3}\right)=\frac{2 \pi}{3} \times 2+\frac{2 \pi^{2}}{9} \times \sqrt{3}=\frac{2 \pi^{2}}{3 \sqrt{3}}+\frac{4 \pi}{3}\)