Solution of the differential equation 2y sin x dy-dx - 2 sin x cos x - y2 cos x satisfying y -2 - 1is given by

The given equation can be written as 2y sin x dy/dx + y^2 cos x = sin 2x ⇒d/dx (y^2 sin x) = sin 2x ⇒ y^2 sin x = ( 1/2) cos 2x + C. So (y (π/2))^2 sin ...

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Standard XII

Mathematics

Particular Solution of a Differential Equation

Solution of t...

Question

Solution of the differential equation $2 y s i n x \frac{d y}{d x} = 2 s i n x c o s x - y^{2} c o s x$ satisfying $y (\frac{π}{2}) = 1)$ is given by

y^{2} = s i n x

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y = s i n^{2} x

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y^{2} = c o s x + 1

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y^{2} = s i n x = 4 c o s^{2} x

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Solution

The correct option is A $y^{2} = s i n x$
The given equation can be written as $2 y s i n x \frac{d y}{d x} + y^{2} c o s x = s i n 2 x$
$\Rightarrow \frac{d}{d x} (y^{2} s i n x) = s i n 2 x \Rightarrow y^{2} s i n x = (- \frac{1}{2}) c o s 2 x + C .$
So $(y (\frac{π}{2}))^{2} s i n (\frac{π}{2}) = (- \frac{1}{2}) c o s (\frac{2 π}{2}) + C \Rightarrow C = \frac{1}{2}$
Hence $y^{2} s i n x = (\frac{1}{2}) (1 - c o s 2 x) = s i n^{2} x \Rightarrow y^{2} = s i n x$

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Solution of the differential equation 2y sin xdydx=2 sin x cos x−y2 cos x satisfying y(π2)=1)is given by

The correct option is A y2=sin xThe given equation can be written as 2y sin xdydx+y2 cos x=sin 2x ⇒ddx(y2 sin x)=sin 2x⇒y2 sin x=(−12)cos 2x+C. So (y(π2))2 sin(π2)=(−12)cos(2π2)+C⇒C=12 Hence y2sinx=(12)(1−cos2x)=sin2x⇒y2=sinx

Solution of the differential equation $2 y s i n x \frac{d y}{d x} = 2 s i n x c o s x - y^{2} c o s x$ satisfying $y (\frac{π}{2}) = 1)$ is given by