Question

The solution of the differential equation $\frac{dy}{dx}=\frac{y^2-y-2}{x^2+2x-3}$ is

• A. $\frac{1}{3}\ln \left|\frac{y-2}{y+1}\right|=\frac{1}{2}\ell n\left|\frac{x+3}{x-1}\right|+c$
• B. $\frac{1}{3}\ell n\left|\frac{y+1}{y-2}\right|=\frac{1}{4}\ln \left|\frac{x-1}{x+3}\right|+c$
• C. $2\ell n\left|\frac{y+1}{y-2}\right|=3\ell n\left|\frac{x+3}{x-1}\right|+c$
• D. $4\ln \left|\frac{y-2}{y+1}\right|=3\ln \left|\frac{x-1}{x+3}\right|+c$

Answer 1

Answer: (D)

$\frac{dy}{dx}=\frac{y^2-y-2}{x^2+2x-3}$
$\frac{dy}{(y-2)(y+1)}=\frac{dy}{(x+3)(x-1)}$
$\Rightarrow \int \frac{dy}{(y-2)(y+1)}=\int \frac{dy}{(x+3)(x-1)}$
$\Rightarrow \frac{1}{3}\int \left(\frac{1}{y-2}-\frac{1}{y+1}\right)dy=\frac{1}{4}\int \left(\frac{1}{x-1}-\frac{1}{x+3}\right)dx$
$\Rightarrow \frac{1}{3}\ln \left|\frac{y-2}{y+1}\right|=\frac{1}{4}\ln \left|\frac{x-1}{x+3}\right|+c$