The general solution of the differential equation \( x(x^{2}-1)\displaystyle\frac{dy}{dx}=1 \) is :
[Where \( C \) is an arbitrary constant]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y=\displaystyle\frac{1}{2}\log_{e}\left|\frac{x^{2}-1}{x^{2}}\right|+C \)
The variables separate at once:
\( \displaystyle dy=\frac{dx}{x(x^2-1)} \)
\( \displaystyle =\frac{dx}{x(x-1)(x+1)} \)
In partial fractions,
\( \displaystyle \frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{D}{x+1} \)
\( \displaystyle x=0:\ A=\frac{1}{(-1)(1)}=-1 \)
\( \displaystyle x=1:\ B=\frac{1}{(1)(2)}=\frac{1}{2} \)
\( \displaystyle x=-1:\ D=\frac{1}{(-1)(-2)}=\frac{1}{2} \)
Integrating both sides:
\( \displaystyle y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+C \)
\( \displaystyle y=\frac{1}{2}\log|x^2-1|-\frac{1}{2}\log x^2+C \)
\( \displaystyle =\frac{1}{2}\log\left|\frac{x^2-1}{x^2}\right|+C \)
Shortcut: Differentiate Option (1), written as \( y=\frac{1}{2}\log|x^2-1|-\log|x| \):
\( \displaystyle \frac{dy}{dx}=\frac{x}{x^2-1}-\frac{1}{x} \)
\( \displaystyle =\frac{x^2-(x^2-1)}{x(x^2-1)} \)
\( \displaystyle =\frac{1}{x(x^2-1)} \)
which is exactly the given equation.