Solving ode form of $\frac{dy}{dx} = \frac{cx+dy}{ax+by}$. \begin{align}
\frac{dy}{dx} = \frac{cx+dy}{ax+by}
\end{align}
i want to solve this differential equation, in two ways. 
First with condition of $ad-bc\neq0$ and without this condition. 
I wonder this condition is really necessary. 

Following mea43, I have 
\begin{align}
&x \frac{dv}{dx} = \frac{c+(d-a)v - bv^2}{a+bv} \\
&\int \frac{1}{x} dx = \int \frac{a+bv}{c+(d-a)v - bv^2} dv =
-\frac{(a+d) \text{ArcTan}\left[\frac{a-d+2 b v}{\sqrt{-a^2-4 b c+2 a d-d^2}}\right]}{\sqrt{-a^2-4 b c+2 a d-d^2}}-\frac{1}{2} \text{Log}[-c+v (a-d+b v)]
\end{align}
where i borrow the results of mathematica. Is there some ways to integrate this? 

If $ad=bc$ then $$\frac{a}{c}=\frac{b}{d}=k$$(let us say that ratio is k )
Then $$\frac{dy}{dx}=\frac{cx+dy}{ax+by}=\frac{1}{k}\frac{ax+by}{ax+by}=\frac{1}{k}$$
Hence $y=\frac{x}{k}+c$
But if $ad-bc\ne 0$
Let $y=vx$ then $\frac{dy}{dx}=v+x\frac{dv}{dx}$.
Hence $$ \frac{dy}{dx}=\frac{cx+dy}{ax+by}$$
$$\Rightarrow v+x\frac{dv}{dx}=\frac{c+dv}{a+bv}$$ we get an  entirely different solution
