Solving a seperable ODE using substitution proof This is the question my softmore differential equations professor asked on our practice exam: 

Show that the substitution $y=ux$ in the first-order differential equation
  $$p(x,y)\;dx+q(x,y)\;dy=0$$
  results in an ODE (in u and x) which can be solved by "separation of variables" if p and q are homogeneous of the same degree (meaning $p(tx,ty)=t^dp(x,y)$ and $q(yx,ty)=t^dp(x,y)$ for some integer d and all real $t\neq0$.)

What is the expression you get for the solution of the original ODE (i.e., after undoing the substitution). Check that it works.

Note: you'll need to have the correct substitution for dy here for things to work.



This is how far I've gotten

Where should I go from here? I am completely lost. HELP!

Unfortunately I did not manage to read your notes. Follow a deduction in case it can help.
Making $y = \lambda(x) x$ and considering $dy = \lambda dx+x d\lambda$ we have
$$
p(x,\lambda x)dx+q(x,\lambda x ) dy = x^qp(1,\lambda)dx + x^q q(1,\lambda)(\lambda dx+x d\lambda) = 0
$$
so we follow with
$$
p(1,\lambda)dx + q(1,\lambda)(\lambda dx+x d\lambda) = (p(1,\lambda)+\lambda q(1,\lambda))dx + x q(1,\lambda)d\lambda=0
$$
hence
$$
\frac{q(1,\lambda)d\lambda}{p(1,\lambda)+\lambda q(1,\lambda)} + \frac{dx}{x}=0
$$
