Differential Equations Question 122
Question: The equation of the curve through the point (1,0) and whose slope is $ \frac{y-1}{x^{2}+x} $ is
Options:
A) $ (y-1)(x+1)+2x=0 $
B) $ 2x(y-1)+x+1=0 $
C) $ x(y-1)(x+1)+2=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Slope $ =\frac{dy}{dx} $
Therefore $ \frac{dy}{dx}=\frac{y-1}{x^{2}+x} $
Therefore $ \frac{dy}{y-1}=\frac{dx}{x^{2}+x} $
Therefore $ \int _{{}}^{{}}{\frac{1}{y-1}}dy=\int _{{}}^{{}}{( \frac{1}{x}-\frac{1}{x+1} )}dx+c $
Therefore $ \frac{(y-1)(x+1)}{x}=k $
Putting $ x=1 $ , $ y=0 $ , we get $ k=-2 $
Hence the equation is $ (y-1)(x+1)+2x=0 $ .
BETA
