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Variable Separable Equation with Partial Fractions | JEE

JEE Maths question with a full step-by-step solution.

Question

The solution of

\dfrac{dy}{dx} = \dfrac{y^2 - y - 2}{x^2 + 2x - 3}

is (where

c

is an arbitrary constant)

\dfrac{1}{3}\log\left(\dfrac{y-2}{y+1}\right) = \dfrac{1}{4}\log\left(\dfrac{x+3}{x-1}\right) + c

\dfrac{1}{3}\log\left(\dfrac{y+1}{y-2}\right) = \dfrac{1}{4}\log\left(\dfrac{x-1}{x+3}\right) + c

4\log\left(\dfrac{y-2}{y+1}\right) = 3\log\left(\dfrac{x-1}{x+3}\right) + c

correct

DNone of these

Solution

Step 1: Separate and factor:

\frac{dy}{(y-2)(y+1)} = \frac{dx}{(x+3)(x-1)}.

Step 2: Partial fractions:

\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:

\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

12

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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