Differential Equations
Particular Solution
MMTS_Full_Test_08
Grade 12
Question:
The solution curve of $\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}$, $xy\ne 1$, passing through origin is
$y=x$
$y=-x$
$y=1/(x+1)$
$y=x/(1+x)$
Step-by-Step Solution
Key Concept: Separate variables: $\frac{dy}{y^2-1}=\frac{dx}{x^2-1}$; integrate using partial fractions
$\ln\frac{y-1}{y+1}=\ln\frac{x-1}{x+1}+C$. At $(0,0)$: $\ln\frac{-1}{1}=\ln\frac{-1}{1}+C\Rightarrow C=0$. $\frac{y-1}{y+1}=\frac{x-1}{x+1}\Rightarrow y=x$.
Correct Answer: 1