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Mathematics Question on Differential equations

If (2+sin⁡ x)dydx+(y+1)cos⁡ x=0(2 + \sin \, x ) \frac{dy}{dx} + (y + 1) \cos \, x = 0 and y(0)=1,y(0) = 1, then y(π2)y \left( \frac{\pi}{2} \right) is equal to :

A

−23- \frac{2}{3}

B

−13- \frac{1}{3}

C

43 \frac{4}{3}

D

13 \frac{1}{3}

Answer

13 \frac{1}{3}

Explanation

Solution

(2+sin⁡x)dydx+(y+1)cos⁡x=0\left(2+\sin x\right) \frac{dy}{dx}+\left(y+1\right) \cos x = 0
y(0)=1,y(π2)y (0) = 1, y \left(\frac{\pi}{2}\right) = ?
1y+1dy+cos⁡x2+sin⁡xdx=0\frac{1}{y+1}dy+\frac{\cos x}{2+\sin x} dx = 0
In∣y+1∣+In(2+sin⁡x)=InCIn\left|y+1\right|+In\left(2+\sin x\right) = InC
(y+1)(2+sin⁡x)=C(y + 1) (2 + \sin x) = C
Put x=0, y=1x = 0,\, y = 1
(1+1)⋅2=C⇒C=4(1 + 1) \cdot 2 = C \Rightarrow C = 4
Now, (y+1)(2+sin⁡x)=4(y +1)(2 + \sin x) = 4
For, x=π2x = \frac{\pi}{2}
(y+1)(2+1)=4(y +1)(2 +1) = 4
y+1=43y + 1 = \frac{4}{3}
y=43−1=13y = \frac{4}{3}-1 =\frac{1}{3}