Mathematics - Differential Equation Question with Solution | TestHub

Given, $\frac{dy}{dx}-3y\cot x=\sin 2x,y\left(\frac{\pi }{2}\right)=2$ [Linear differential equation]\ ∴ I.F. $={\mathrm{e}}^{-\int 3\cot \mathrm{xdx}}={\mathrm{e}}^{-3\ln (\sin \mathrm{x})}=\frac{1}{{\sin }^3\mathrm{x}}$ ∴ General solution is $\mathrm{y}\left(\frac{1}{{\sin }^3\mathrm{x}}\right)=\int \frac{2\sin \mathrm{x}\cdot \cos \mathrm{x}}{{\sin }^3\mathrm{x}}\mathrm{dx}+\mathrm{C}=2\int \cos \mathrm{ecx}\cot \mathrm{x}\mathrm{dx}+\mathrm{C}$. $\Rightarrow \frac{y}{{\sin }^{3}x}=-2\cos ecx+C$ As, $y\left(x=\frac{\pi }{2}\right)=2\Rightarrow \frac{2}{1^3}=-2+C\Rightarrow C=4\therefore y=\left(4{\sin }^{3}x-2{\sin }^{2}x\right)$

(A) $\mathrm{y}\left(\frac{\pi }{6}\right)=4{\left(\frac{1}{2}\right)}^3-2{\left(\frac{1}{2}\right)}^2=\frac{1}{2}-\frac{1}{2}=0$