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Variable Separable Equation with Partial Fractions | JEE

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Question
The solution of dydx=y2−y−2x2+2x−3\dfrac{dy}{dx} = \dfrac{y^2 - y - 2}{x^2 + 2x - 3} is (where cc is an arbitrary constant)
A13log⁡(y−2y+1)=14log⁡(x+3x−1)+c\dfrac{1}{3}\log\left(\dfrac{y-2}{y+1}\right) = \dfrac{1}{4}\log\left(\dfrac{x+3}{x-1}\right) + c
B13log⁡(y+1y−2)=14log⁡(x−1x+3)+c\dfrac{1}{3}\log\left(\dfrac{y+1}{y-2}\right) = \dfrac{1}{4}\log\left(\dfrac{x-1}{x+3}\right) + c
C4log⁡(y−2y+1)=3log⁡(x−1x+3)+c4\log\left(\dfrac{y-2}{y+1}\right) = 3\log\left(\dfrac{x-1}{x+3}\right) + ccorrect
DNone of these
Solution
Step 1: Separate and factor:
dy(y−2)(y+1)=dx(x+3)(x−1).\frac{dy}{(y-2)(y+1)} = \frac{dx}{(x+3)(x-1)}.
 Step 2: Partial fractions:
1(y−2)(y+1)=13(1y−2−1y+1),1(x+3)(x−1)=14(1x−1−1x+3).\frac{1}{(y-2)(y+1)} = \frac{1}{3}\left(\frac{1}{y-2} - \frac{1}{y+1}\right), \qquad \frac{1}{(x+3)(x-1)} = \frac{1}{4}\left(\frac{1}{x-1} - \frac{1}{x+3}\right).
 Step 3: Integrate:
13log⁡∣y−2y+1∣=14log⁡∣x−1x+3∣+c′.\frac{1}{3}\log\left|\frac{y-2}{y+1}\right| = \frac{1}{4}\log\left|\frac{x-1}{x+3}\right| + c'.
 Multiplying through by 1212:
4log⁡(y−2y+1)=3log⁡(x−1x+3)+c.4\log\left(\frac{y-2}{y+1}\right) = 3\log\left(\frac{x-1}{x+3}\right) + c.
 Correct answer: (3)
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