Find the general solution of the Euler homogeneous equations?
dy/dx= (2y-x)/(2x-y)
Using the substitution y=vx
v+x(dv/dx)= (2xv-x)/(2x-xv)
v+ x(dv/dx)= x(2v-1)/x(2-v)
v+ x(dv/dx)= (2v-1)/(2-v)
x(dv/dx)= ((2v-1)/(2-v))-v
x(dv/dx)= (v^2-1)/(2-v)
integral ((2-v)/(v^2-1))dv = integral (1/x)dx
integral (2-v)/((v-1)*(v+1))dv = integral (1/x)dx
integral 1/(2v-2)dv - integral 3/(2v+2)dv = integral 1/x dx
lnx + c = 0.5ln(2v-2)- 1.5ln(2v+2)
lnx + c = 0.5ln((2y/x)-2)- 1.5ln((2y/x)+2)
Is this right so far and what do I do now?
You have some slight issues from the point where I start. We have:
$$\displaystyle \int \dfrac{2-v}{v^2-1}~dv = \int \dfrac 1x ~dx$$
This yields:
$$\dfrac 12 \ln (1 - v) - \dfrac 32 \ln (1 + v) = \ln x + c$$
Now, we substitute back $v = \dfrac yx$, yielding:
$$\dfrac 12 \ln \left(1 - \dfrac yx\right) - \dfrac 32 \ln \left(1 + \dfrac yx\right) = \ln x + c$$
Hopefully, you see the slight algebra issue. Also, it does not look like we can simplify this to isolate $y(x)$ to anything useful, so we are done.