First slide

Methods of solving first order first degree differential equations

Question

$Let y = y (x) be the solution of the differential equation \sin x \frac{d y}{d x} + y \cos x = 4 x, \in (0, π) . If$ $y (\frac{π}{2}) = 0, then y (\frac{π}{6}) is equal to$

Easy

A

$- \frac{4}{9} π^{2}$
B

$\frac{4}{9 \sqrt{3}} π^{2}$
C

$\frac{- 8}{9 \sqrt{3}} π^{2}$
D

$- \frac{8}{9} π^{2}$

Solution

$Given that \sin x \frac{d y}{d x} + y \cos x = 4 x$
$\Rightarrow \frac{d}{d x} (y \sin x) = 4 x$
Integration on both sides
$\int d (y \sin x) = \int 4 x d x$
$\Rightarrow y \sin x = 4 \frac{x^{2}}{2} + C$
$\Rightarrow y \sin x = 2 x^{2} + C$
$Given y (\frac{π}{2}) = 0$
$C = - \frac{π^{2}}{2}$
$Thus y \sin x = 2 x^{2} - \frac{π^{2}}{2}$
$Now y \sin \frac{π}{6} = 2 (\frac{π^{2}}{36}) - \frac{π^{2}}{2}$
$\Rightarrow y \cdot \frac{1}{2} = \frac{π^{2}}{18} - \frac{π^{2}}{2}$
$\Rightarrow \frac{y}{2} = - \frac{{8π}^{2}}{18}$
$∴ y = - \frac{{8π}^{2}}{9}$

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