Answer
\[y(x)=C+C_{1}\ln\left|\frac{x-a}{x-b}\right|^\frac{1}{(a-b)}\]
Work Step by Step
$$(x-a)(x-b)y'-(y-c)=0$$
$$y'=\frac{dy}{dx}=\frac{y-c}{(x-a)(x-b)}$$
Separating variables
\[\frac{dy}{y-c}=\frac{dx}{(x-a)(x-b)}\]
Integrating,
\[\int\frac{dy}{y-c}=\int\frac{dx}{(x-a)(x-b)}+C_{1}\]
Where $C_{1}$ is constant of integration
$\int\frac{dy}{y-c}=\frac{1}{(a-b)}\int\left(\frac{1}{x-a}-\frac{1}{x-b}\right)dx+C_{1}$
$\ln|y-c|=\frac{1}{(a-b)}[\ln|x-a|-\ln|x-b|]+C_{1}$
\[\ln|y-c|=\frac{1}{(a-b)}\ln\left|\frac{x-a}{x-b}\right|+C_{1}\]
\[y(x)=C+C_{1}\ln\left|\frac{x-a}{x-b}\right|^\frac{1}{(a-b)}\]
