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If $\left ( 2+\sin x \right )\: \frac{dy}{dx}\: +\left ( y+1 \right )\cos x= 0$

and y(0)=1,then $y\left ( \frac{\pi }{2} \right )$ is equal to :

Option 1)
$-\frac{2}{3}$

Option 2)
$-\frac{1}{3}$

Option 3)
$\frac{4}{3}$

Option 4)
$\frac{1}{3}$

Answers (1)

best_answer

As we learnt in

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

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

$\left ( 2+\sin x \right )\frac{dy}{dx}+\left ( y+1 \right )\cos x=0$

$=>\frac{dy}{dx}=-\left ( y+1 \right ).\frac{\cos x}{2+\sin x}$

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

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

$=>-log(y+1)=log(2+\sin x)+C$

$=> put \: x=0,\: y=1$

$-log2=log(2+\sin x)+C=+log2+C$

$\therefore C=-2log2$ $=log\frac{1}{4}$

$\therefore -log(y+1)=log(2+\sin x)+log\frac{1}{4}$

Now put

$x=\frac{\pi }{2}$

$\therefore -log(y+1)=log(2+1)+log\frac{1}{4}$ $=log\frac{3}{4}$

$\therefore y+1=\frac{4}{3}$

$=> y=\frac{1}{3}$

Option 1)

$-\frac{2}{3}$

Incorrect option

Option 2)

$-\frac{1}{3}$

Incorrect option

Option 3)

$\frac{4}{3}$

Incorrect option

Option 4)

$\frac{1}{3}$

Correct option

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Aadil

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