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If $y=y(x)$ and $\frac{2+\sin x}{y+1}\frac{dy}{dx}=-\cos x, \: y(0)=1,$ then $y(\frac{\pi }{2})$ is equal to

Option 1)
$\frac{1}{3}$

Option 2)
$\frac{2}{3}$

Option 3)
$-\frac{1}{3}$

Option 4)
1

Answers (1)

best_answer

Using

Solution of Differential Equation -

$\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )$

put

$Z =ax+by+c$

- wherein

Equation with convert to

$\int \frac{dz}{bf\left ( z \right )+a} =x+c$

$\frac{dy}{dx}.\frac{1}{1+y}= -\frac{\cos x}{2+\sin x}$

$\int \frac{dy}{1+y} = \int \frac{-\cos x}{2+\sin x}dx\ \ \ \ Let \ 2\times \sin x=t,\ \ \cos xdx=dt$

$=\int \frac{-dt}{t}= -\log t+C$

$\log \left ( 1+y \right ) = -\log \left ( 2+ \sin x \right )+C\ \ \ \ -(i)$

$\log \left ( 1+1 \right ) = -\log \left ( 2+ 0 \right )+C= -\log 2+C$

$\therefore C=\log 4$

$\therefore \left ( 1+y \right )= \frac{4}{2+\sin x}\ \ \ from \ (i)$

$1+y = \frac{4}{2+1}= \frac{4}{3}\ \ \left [ \because \sin \frac{\pi }{2} =1\right ]$

$y = \frac{4}{3} - 1=\frac{1}{3}$

Option 1)

$\frac{1}{3}$

This option is correct

Option 2)

$\frac{2}{3}$

This option is incorrect

Option 3)

$-\frac{1}{3}$

This option is incorrect

Option 4)

1

This option is incorrect

Posted by

divya.saini

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