Resolver la ecuacion diferencial

Es una ecuación homogénea podemos haber un cambio de variable u=y/x

$$\begin{align}&\frac{dy}{dx}=\frac{5x+4y}{2x-y}=\frac{5+4 \frac yx}{2-\frac yx}\\ &\\ &u= \frac yx \quad y=ux\\ &\\ &\frac{dy}{dx} = \frac{du}{dx}x+u\\ &\\ &\\ &\frac{du}{dx}x+u=\frac{5+4u}{2-u}\\ &\\ &\frac{du}{dx}x=\frac{5+4u}{2-u}-u\\ &\\ &\frac{du}{dx}x = \frac{5+4u-2u+u^2}{2-u}\\ &\\ &\frac{du}{dx}x = \frac{u^2+2u+5}{2-u}\\ &\\ &\left(\frac{2-u}{u^2+2u+5}  \right)du=\frac {dx}{x}\\ &\\ &\\ &-\frac 12\left(\frac{2u-4}{u^2+2u+5}  \right)du=\frac{dx}{x}\\ &\\ &-\frac 12 \left(\frac{2u+2}{u^2+2u+5}-\frac{6}{u^2+2u+5}  \right)du=\frac{dx}{x}\\ &\\ &-\frac 12 ln(u^2+2u+5)+3\int \frac{du}{u^2+2u+5} = lnx+ C\\ &\\ &\text{hacemos la integral aparte}\\ &\\ &\int \frac{du}{(u+1)^2+4}=\frac 14\int \frac{du}{\left(\frac{u+1}{2}\right)^2+1}=\\ &\\ &\frac 12 arctg \left(\frac{u+1}{2}\right)\\ &\\ &\text{Y la solución es}\\ &\\ &-\frac 12 ln(u^2+2u+5)+\frac 32 arctg \left(\frac{u+1}{2}\right) = lnx+ C\\ &\\ &-\frac 12 ln\left(\frac{y^2}{x^2}+\frac{2y}x+5\right)+\frac 32 arctg \left(\frac{\frac yx+1}{2}\right) = lnx+ C\\ &\\ &\\ &-\frac 12 ln\left(\frac{y^2}{x^2}+\frac{2y}x+5\right)+\frac 32 arctg \left(\frac{y+x}{2x}\right) = lnx+ C\end{align}$$

Y eso es todo.