Question

$\frac{dy}{dx}=\frac{y^2-2y}{x-1}$

Answer 1

Consider given the deferential equation,

$\frac{dy}{dx}=\frac{y^2-2y}{x-1}$

$\frac{1}{y^2-2y}dy=\frac{1}{x-1}dx$

Taking integration both sides,

$\int \frac{1}{y^2-2y}dy=\int \frac{1}{x-1}dx$

$\int \frac{1}{y(y-2)}dy=\int \frac{1}{x-1}dx$

$\int [\frac{1}{y-2}-\frac{1}{y}]dy=\int \frac{1}{x-1}dx$

$\int \frac{1}{y-2}dy-\int \frac{1}{y}dy=\int \frac{1}{x-1}dx$

$\log (y-2)-\log y=\log (x-1)+\log C$

$lof\frac{y-2}{y}=\log \frac{(x-1)}{C}$

$\frac{y-2}{y}=\frac{x-1}{C}$

$C(y-2)=y(x-1)$

Hence, this is the answer.