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Variable Separable Differential Equation – Solved | JEE

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Question
If y=y(x)y = y(x) and (2+sin⁡xy+1)dydx=−cos⁡x\left(\dfrac{2 + \sin x}{y + 1}\right)\dfrac{dy}{dx} = -\cos x with y(0)=0y(0) = 0, then y(5π6)y\left(\dfrac{5\pi}{6}\right) equals:
A15\dfrac{1}{5}
B−15-\dfrac{1}{5}correct
C35\dfrac{3}{5}
D25\dfrac{2}{5}
Solution
Step 1: Separate the variables:
dyy+1=−cos⁡x2+sin⁡x dx.\frac{dy}{y + 1} = \frac{-\cos x}{2 + \sin x}\,dx.
 Step 2: Integrate (the right side is −dln⁡(2+sin⁡x)-d\ln(2 + \sin x)):
ln⁡(y+1)=−ln⁡(2+sin⁡x)+ln⁡c⇒y+1=c2+sin⁡x.\ln(y + 1) = -\ln(2 + \sin x) + \ln c \Rightarrow y + 1 = \frac{c}{2 + \sin x}.
 Step 3: Apply y(0)=0y(0) = 0: 1=c2⇒c=21 = \dfrac{c}{2} \Rightarrow c = 2, so y+1=22+sin⁡xy + 1 = \dfrac{2}{2 + \sin x}. Step 4: At x=5π6x = \dfrac{5\pi}{6}, sin⁡5π6=12\sin\dfrac{5\pi}{6} = \dfrac{1}{2}:
y+1=22+12=45⇒y=−15.y + 1 = \frac{2}{2 + \frac{1}{2}} = \frac{4}{5} \Rightarrow y = -\frac{1}{5}.
 Correct answer: (2)
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