Question

The solution of the differential equation:
$(x^2-1)\frac{dy}{dx}+2xy=\frac{1}{(x^2-1)}$

• A. $y(x^2-1)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C$
• B. $y(x^2+1)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|-C$
• C. $y(x^2-1)=\frac{5}{2}\log \left|\frac{x-1}{x+1}\right|+C$
• D. None of these

Answer 1

The correct option is A $y(x^2-1)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C$

$(x^2-1)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$

$\Rightarrow$ $\frac{d}{dx}\left[y\right(x^2-1\left)\right]=\frac{1}{x^2-1}$

$\Rightarrow$ $y(x^2-1)=\int \frac{dx}{x^2-1^2}\Rightarrow y(x^2-1)=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+c$