Mathematics - Differential Equation Question with Solution | TestHub

$\frac{dy}{dx}-y\tan x=2x\sec x$ I.F. $=e^{-\int \tan xdx}=\cos x$ $\therefore y\cdot \cos x=\int 2xdx=x^2+c$ Now, $y(0)=0\Rightarrow c=0,\therefore y=x^2\sec x$ $\Rightarrow y^{\prime }=2x\sec x+x^2\sec x\tan x$ Now, $y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}\times \sqrt{2}=\frac{{\pi }^2}{8\sqrt{2}}$ $y\left(\frac{\pi }{3}\right)=\frac{{\pi }^2}{9}\times 2=\frac{2{\pi }^2}{9}$ $y^{\prime }\left(\frac{\pi }{4}\right)=\frac{2\pi }{4}\times \sqrt{2}+\frac{{\pi }^2}{8\sqrt{2}}\times 1=\frac{{\pi }^2}{8\sqrt{2}}+\frac{\pi }{\sqrt{2}}$ $y^{\prime }\left(\frac{\pi }{3}\right)=\frac{2\pi }{3}\times 2+\frac{2{\pi }^2}{9}\times \sqrt{3}=\frac{2{\pi }^2}{3\sqrt{3}}+\frac{4\pi }{3}$