MCQ
A function $y = f(x)$ satisfies the differential equation $f(x).sin\ 2x\ -\ cos\ x\ +\ (1 + sin^2x) f'(x) = 0$ where $f(0) = 0$ . Then value of $f(\frac {\pi}{6})$ is equal to
- A$\frac {1}{5}$
- B$\frac {3}{5}$
- C$\frac {4}{5}$
- ✓$\frac {2}{5}$
where $y=f(x)$
$\frac{d y}{d x}+\left(\frac{\sin 2 x}{1+\sin ^{2} x}\right) y=\frac{\cos x}{1+\sin ^{2} x}$
I.F. $=e^{\int \frac{\sin 2 x}{1+\sin ^{2} x} d x}=e^{\ln \left(1+\sin ^{2} x\right)}=1+\sin ^{2} x$
$\mathrm{y}\left(1+\sin ^{2} \mathrm{x}\right)=\sin \mathrm{x}+\mathrm{C} ; \quad(\mathrm{y}(0)=0)$
$\Rightarrow \mathrm{C}=0$
hence, $y=\frac{\sin x}{1+\sin ^{2} x} \quad y\left(\frac{\pi}{6}\right)=\frac{2}{5}$
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