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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

Text

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

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

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

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  • $\begingroup$ This is indeed where I ended up... I have no idea how to integrate from here though $\endgroup$
    – Phil
    May 5 '20 at 21:34

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