Showing $\lim_{t \rightarrow T} [X(t) + X'(t)] < \infty$ for $X''(t) + X'(t) + X(t) = 0$.

Question

Suppose $X \in C^2 [0,T]$ solves $$X''(t) + X'(t) + X(t) = 0$$ with the initial conditions as $X(0)=1, X'(0) = 0$.

I was trying to show that $\lim_{t \rightarrow T} [X(t) + X'(t)] < \infty$ and also how the finiteness of the above limit could assure that the solution is global?

I thought of multiplying $X'$ on both sides of the differential equation which gives us - $X'X'' + (X')^2 + XX' = 0$ and then integrating both sides from $0$ to $t$ giving us -

$\int_{0}^{t} X'X'' + \int_{0}^{t} (X')^2 dt + \int_{0}^{t} X dt = t$

I was guessing that may be a product rule could simplify a bit but it becomes a bit complex after this.

Like $(X'X)' = X''X + (X')^2$

$(X'^2)' = 2X''X'$

Any ideas?

Answer 1

This is a second order autonomous differential equation, and with techniques in the link, it's pretty simple to exactly find an implicit solution in terms of integration.