Solution of 2ysin xdydx-2sin x-cos x-y2cos x-x-2-y-1 is given by

2ysin xdydx=2sin x·cos x y^2cos x ⇒2ysin xdydx+y^2cos x=2sin x·cos x ⇒sin xddx(y^2)+y^2ddxsin x=2sin x·cos x ⇒ddx(y^2sin x)=2sin x·cos x ⇒∫ d(y^2sin x)=∫sin2x· ...

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Standard XIII

Mathematics

General Solution of a Differential Equation

Solution of 2...

Question

Solution of $2 y sin x \frac{d y}{d x} = 2 sin x \cdot cos x - y^{2} cos x, x = \frac{π}{2}, y = 1$ is given by

y^{2} = sin x

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y = {sin}^{2} x

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y^{2} = cos x + 1

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None of these

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Solution

The correct option is A $y^{2} = sin x$
$2 y sin x \frac{d y}{d x} = 2 sin x \cdot cos x - y^{2} cos x$
$\Rightarrow 2 y sin x \frac{d y}{d x} + y^{2} cos x = 2 sin x \cdot cos x$
$\Rightarrow sin x \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} sin x = 2 sin x \cdot cos x$
$\Rightarrow \frac{d}{d x} (y^{2} sin x) = 2 sin x \cdot cos x$
$\Rightarrow \int d (y^{2} sin x) = \int sin 2 x \cdot d x$
$\Rightarrow y^{2} sin x = - \frac{cos 2 x}{2} + c$
At $x = \frac{π}{2}, y = 1$
$1^{2} \cdot sin \frac{π}{2} = - \frac{cos π}{2} + c \Rightarrow c = \frac{1}{2}$
So, $y^{2} sin x = - \frac{cos 2 x}{2} + \frac{1}{2}$
$\Rightarrow y^{2} = \frac{1 - cos 2 x}{2 sin x}$
$\Rightarrow y^{2} = sin x$

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Solution of 2ysinxdydx=2sinx⋅cosx−y2cosx,x=π2,y=1 is given by

Solution of $2 y sin x \frac{d y}{d x} = 2 sin x \cdot cos x - y^{2} cos x, x = \frac{π}{2}, y = 1$ is given by