Let y=vx, dy/dx=v+xdv/dx.
The DE becomes:
(x-vx)(4x+vx)+(5x-vx)dy/dx=0,
(1-v)(4+v)+(5-v)(v+xdv/dx)=0,
4-3v-v2+5v+5xdv/dx-v2-vxdv/dx=0, 4+2v-2v2+5xdv/dx-vxdv/dx=0,
x(dv/dx)(5-v)=2v2-2v-4=2(v2-v-2).
This can be written:
[(5-v)/(v2-v-2)]dv=2dx/x and the whole expression is now integrable.
(5-v)/(v2-v-2) can be split into partial fractions:
(5-v)/[(v+1)(v-2)]≡A/(v+1)+B/(v-2),
5-v≡A(v-2)+B(v+1)=Av-2A+Bv+B, so: A+B=-1, -2A+B=5, 3A=-6, A=-2, B=1:
∫[-2/(v+1)+1/(v-2)]dv=2∫dx/x=ln(ax2), where a is a constant.
-ln((v+1)2)+ln|v-2|=ln(ax2), (v-2)/(v+1)2=ax2, ((y/x)-2)/((y/x)+1)2=ax2,
[(y-2x)/x][x2/(x2+y2)]=ax2, (y-2x)/(x2+y2)=ax.