# show locally lipschitz and solve ODE

I have the following initial value problem: $$\dot y =\frac{1}{2}\left\vert \frac {x}{y}+\frac {y^3}{x^3}\right\vert$$ with the initial condition $y(x_0)=y_0$

First i have to show that there exists a unique maximal solution for that I tried to prove that the function is locally Lipschitz with respect to y.

So I checked if $\left\vert f(x,y_1)-f(x,y_2)\right\vert \le L \left\vert y_1-y_2 \right\vert$

But I only got $\vert f(x,y_1)-f(x,y_2)\vert = \frac{1}{2} \left\vert \frac{x^4(y_2-y_1)+y_1y_2(y_1^3-y_2^3)}{x^3y_1y_2} \right\vert$

Now I don´t really now how to get my L here. Apart from that I need to find the general solution for the ODE afterwards and I don´t have a clue about that right now... Thanks for your help

Hint: use the triangle inequality and try to prove the Lipschitz property for the two terms in the resulting sum. For this, you could use the MVT with respect to $y$ (and you should also keep away from some values of $x_0$ and $y_0$).
• Ok hm, all results I get depend on $y_1$ and $y_2$ so its not enough for Lipschitz property, is it? – L.muc Jun 1 '16 at 12:15
• A standard technique is to bound the absolute value of $\partial f/ \partial y$ and to apply the MVT. You should try. – Siminore Jun 1 '16 at 12:20