Let y = y(x) be the solution of the differential equation . If is equal to:

Engineering

Mathematics

Linear Differential Equation

Question

Let y = y(x) be the solution of the differential equation $sinx \frac{dy}{dx} + ycosx = 4 x, x \in (0, π)$ . If $y (\frac{π}{2}) = 0, then y (\frac{π}{6})$ is equal to:

$- \frac{8}{9} π^{2}$

$\frac{4}{9 \sqrt{3}} π^{2}$

$- \frac{4}{9} π^{2}$

$\frac{- 8}{9 \sqrt{3}} π^{2}$

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sin x $\frac{dy}{dx}$ + y cos x = 4x

⇒ $\frac{dy}{dx}$ + y cot x = 4x cosec x

I.F. = $e^{\int cotxdx}$ = e^{ln sin x} = sin x

$∴$ solution is

y × sin x = $\int$ 4x dx = 2x² + C

$∴ x = \frac{π}{2}$ , y = 0 ⇒ C = – $\frac{π^{2}}{2}$

$∴$ y sin x = 2x² – $\frac{π^{2}}{2}$

At x = $\frac{π}{6}$

$\frac{y}{2} = \frac{π^{2}}{18} - \frac{π^{2}}{2} \Rightarrow y = \frac{- 8}{9} π^{2} .$