Question

The solution of the differential equation $\frac{d y}{d x}=\frac{y^2-y-2}{x^2+2 x-3}$ 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)

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