A function y= f(x) satisfies the differential equation $f (x) sin 2 x - cos x + (1 + {sin}^{2} x) f (x) = 0$ with initial condition y(0) = 0. The value of $f (\frac{π}{6})$ is equal to:

\frac{\sqrt{3}}{2 \sqrt{3} - 5}

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\frac{2 \sqrt{3}}{2 \sqrt{3} + 5}

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\frac{\sqrt{3}}{\sqrt{3} - 5}

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\frac{2 \sqrt{3}}{\sqrt{3} - 5}

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Solution

The correct option is C $\frac{2 \sqrt{3}}{2 \sqrt{3} + 5}$
Put $x = \frac{π}{6}$ ,we get
$f (\frac{π}{6}) sin (\frac{π}{3}) - cos (\frac{π}{6}) + (1 + (sin (\frac{π}{6})^{2}) f (\frac{π}{6}) = 0$
$f (\frac{π}{6}) = \frac{2 \sqrt{3}}{2 \sqrt{3} + 5}$

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A function y= f(x) satisfies the differential equation f(x)sin2x−cosx+(1+sin2x)f(x)=0 with initial condition y(0) = 0. The value of f(π6) is equal to:

The correct option is C 2√32√3+5Put x=π6,we getf(π6)sin(π3)−cos(π6)+(1+(sin (π6)2)f(π6)=0f(π6)=2√32√3+5

A function y= f(x) satisfies the differential equation $f (x) sin 2 x - cos x + (1 + {sin}^{2} x) f (x) = 0$ with initial condition y(0) = 0. The value of $f (\frac{π}{6})$ is equal to:

The correct option is C $\frac{2 \sqrt{3}}{2 \sqrt{3} + 5}$
Put $x = \frac{π}{6}$ ,we get
$f (\frac{π}{6}) sin (\frac{π}{3}) - cos (\frac{π}{6}) + (1 + (sin (\frac{π}{6})^{2}) f (\frac{π}{6}) = 0$
$f (\frac{π}{6}) = \frac{2 \sqrt{3}}{2 \sqrt{3} + 5}$