# Given a unique local solution of ode, how do I prove a unique global solution?

I have the IVP $$\ddot{x}+p(t)\dot x +q(t)x=r(t)$$ $$x(t_0)=x_0$$ $$\dot x(t_0)=x_1$$ where $p(t),r(t),q(t)$ are continuous functions defined for all t in the real numbers. $t_0,x_0,x_1$ are real numbers as well.

Given that there is a $\delta >0$ and a unique function $x(t)\in C^2([t_0,t_0+\delta])$ that satisfies the the IVP for $t_0\leq t \leq t_0+\delta$, (ie, we have a unique local solution) I want to show that a unique global solution exists.

My idea was that we can say that since the interval is arbitrary, that we have a unique solution in $t_0\leq t \leq t_0+\delta$ for all t. Then we can take the union of the local unique solutions and we have a unique global solution...

I am hesitant because this seems too easy, so I am sure I am missing something.

I have read that given a local solution, one has to show that there is a global lipshitz constant so that the solution doesn't blow up in order to get a global solution.

I guess I was just hoping someone could help me out with the question.

Thanks.

To see what you're missing, consider the d.e. $\dot{y} = y^2$. Again, you always have unique local solutions, but you can't just "put them together" to form a global solution on $\mathbb R$ because they blow up, going to $\pm\infty$ at some finite $t$: the general solution is $y = 1/(c - t)$.
EDIT: You may note that if $R = x^2 + \dot{x}^2$, $$\dot{R} = \left | 2 x \dot{x} + 2 \dot{x}\ddot{x} \right| \le A(t) + B(t) R$$ for some continuous $A$ and $B$, and bounds on $A$ and $B$ in some interval can be used to give you bounds on $R$ there.