# 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!

• You have separated the DE now its integrable May 5 '20 at 21:29
• I admit I am tripped up by notation. I don't know how to integrate $$\frac{p(1,u)}{p(1,u) +uq(1,u)} du$$
– Phil
May 5 '20 at 21:32
• well function of x on one side and of u on the other side its seprated May 5 '20 at 21:36
• It's a function of $u$ so it's integrable....you were done in fact May 5 '20 at 21:37

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

$$p(x,\lambda x)dx+q(x,\lambda x ) dy = x^qp(1,\lambda)dx + x^q q(1,\lambda)(\lambda dx+x d\lambda) = 0$$

$$p(1,\lambda)dx + q(1,\lambda)(\lambda dx+x d\lambda) = (p(1,\lambda)+\lambda q(1,\lambda))dx + x q(1,\lambda)d\lambda=0$$
$$\frac{q(1,\lambda)d\lambda}{p(1,\lambda)+\lambda q(1,\lambda)} + \frac{dx}{x}=0$$