Differential Equations Question 34
Question: The solution of the equation $ \frac{dy}{dx}=\frac{y^{2}-y-2}{x^{2}+2x-3} $ is
Options:
A) $ \frac{1}{3}\log | \frac{y-2}{y+1} |=\frac{1}{4}\log | \frac{x+3}{x-1} |+c $
B) $ \frac{1}{3}\log | \frac{y+1}{y-2} |=\frac{1}{4}\log | \frac{x-1}{x+3} |+c $
C) $ 4\log | \frac{y-2}{y+1} |=3\log | \frac{x-1}{x+3} |+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{dy}{dx}=\frac{y^{2}-y-2}{x^{2}+2x-3} $
Therefore $ \frac{dy}{(y-2)(y+1)}=\frac{dx}{(x+3)(x-1)} $
Therefore $ \int _{{}}^{{}}{\frac{dy}{(y-2)(y+1)}}=\int _{{}}^{{}}{\frac{dx}{(x+3)(x-1)}} $
Therefore $ \frac{1}{3}\int _{{}}^{{}}{( \frac{1}{y-2}-\frac{1}{y+1} )}dy=\frac{1}{4}\int _{{}}^{{}}{( \frac{1}{x-1}-\frac{1}{x+3} )}dx $
Therefore $ \frac{1}{3}\log | \frac{y-2}{y+1} |=\frac{1}{4}| \frac{x-1}{x+3} |+c $ .
BETA
