(x)dx + (y2x )dy = 0
it is homogeneous, so I know you multiply by either 1/x or 1/y, but I cannot really get beyond this. Please show as much work as possible…thanks
3 Answers

x dx + (y – 2x)dy = 0
sub u = y/x
xu = y
u dx + x du = dy
x dx + (y – 2x)dy = 0
x dx + (xu – 2x)(u dx + x du) = 0
x dx + xu² dx + x²u du – 2xu dx – 2x² du = 0
(x + xu² – 2xu)dx + (x²u – 2x²) du = 0
x(1 + u² – 2u)dx + x²(u – 2) du = 0
(1 + u² – 2u)dx = x(2 – u) du
dx/x = (2 – u) du/(u² – 2u + 1)
dx/x = (2 – u) du/(u – 1)²
∫dx/x = ∫(2 – u) du/(u – 1)²
sub v = u – 1
dv = du
u = v + 1
∫dx/x = ∫(2 – (v + 1)) dv/v²
∫dx/x = ∫(1 – v) dv/v²
∫dx/x = ∫(1/v² – 1/v) dv
ln  x  = 1/v – lnv + C
Back substitute
ln  x  = 1/(u + 1) – lnu + 1 + C
ln  x  = 1/(y/x + 1) – lny/x + 1 + C
simplify
ln  x  = x/(y + x) – ln(y + x)/x + C
ln  x  = x/(y + x) – lny + x + lnx + C
0 = x/(y + x) – lny + x + C
x/(y + x) + lny + x = C
Making either x or y the subject is difficult

Substitution? I don’t know what you mean by 1/x or 1/y, but here’s how I would solve it. I’d use a change of variables. Is that what you mean?
There’s a theorem for homogeneous D.E.’s. If M(x,y)dx+N(x,y)dy=0 is homogeneous, then it can be solved by making the substitution y=vx.
xdx+(y2x)dy=0 Use y=vx, and thus dy=xdv+vdx
xdx+(vx2x)(xdv+vdx)=0
xdx+vx^2dv+v^2xdx2x^2dv2vxdx=0 (Check my algebra on that)
(x+v^2x2vx)dx+(vx^22x^2)dv=0 (Move some stuff around…)
x(1+v^22v)dx=x^2(v2)dv (Divide by x)
(1+v^22v)dx=x(v2)dv (finally….)
dx/x=(v2)/(1+v^22v)dv (NOW we can integrate.)
lnx=integral of (v2)/(v^22v+1)dv
I stopped here because I think you can perform the integration. Once you get everything in terms of x and v, then from the first substitution, v=y/x. Then it’s just simplifying. I did this in just a couple minutes, so no guarantee I made no mistakes in the algebra. But the theorem is solid.

Let x = vy. Then dx = vdy + ydv. So
vy(vdy + ydv) + (y – 2vy)dy = 0
v^2ydy + vy^2dv + (y – 2vy)dy = 0
vy^2dv + (y – 2vy + v^2y)dy = 0
vy^2dv + y(1 – 2v + v^2)dy = 0
vdv/(1 – 2v + v^2) + dy/y = 0
vdv/(v – 1)^2+ + dy/y = 0
(v – 1 + 1)dv/(v – 1)^2+ + dy/y = 0
dv/(v – 1) + dv/(v – 1)^2+ + dy/y = 0
lnv – 1 – 1/(v – 1) + lny = C
lnx/y – 1 – 1/(x/y – 1) + lny = C
ln(x – y)/y – 1/(x/y – 1) + lny = C
lnx – y – lny – y/(x – y) + lny = C
lnx – y = C + y/(x – y)
x – y = Ke^[y/(x – y)]