[Math] Solution of Exact and Homogeneous differential equation

Question

Consider the equation
$(5y – 2x) (\dfrac{dy}{dx}) – 2y = 0$
This equation is Exact and Homgeneous differential equation. When i use the exact method then i get the following solution
$2xy -(5/2) y^2 = c$
And when I use the Homogeneous method of solution that is putting $y = vx$ in Homogeneous differential equation then I get the following solution
$ln (y) + \dfrac{2x}{5y} = c$.
Question is that are both right solutions. Im confused about different solution. Explain it please.

Answer 1

$$(5y - 2x) (\dfrac{dy}{dx}) - 2y = 0$$ $$ y' =\frac {2y}{(5y - 2x)}$$ Since the equation is homogenous then substitute $y=tx$ $$ t'x+t =\frac {2t}{(5t - 2)}$$ $$ t'x =\frac {4t-5t^2}{(5t - 2)}$$ $$ \int \frac{(5t - 2)} {4t-5t^2}dt=\ln|x|+K$$ $$ -\frac 12\int \frac 1tdt+\frac 52\int \frac{ dt} {4-5t}=\ln|x|+K$$ $$ -\frac 12\ln|t|-\frac 12\int \frac{ dt} {t-4/5}=\ln|x|+K$$ $$ -\frac 12\ln|t|-\frac 12\ln|t-4/5|=\ln|x|+K$$ $$ \ln|y|+\ln|\frac yx-4/5|+\ln|x|=K$$ $$ xy(\frac yx-\frac 45)=K$$ Exactly what you had with exact differential method $$\boxed{ y^2-\frac {4}5xy=K}$$