Assume the IVP $$ x'''+\frac{1}{1+t^2}x''+\sin t \,x'+\frac{t}{x^2+y^2+1}=0 $$ $$ y''+e^{-t}y'+\cos (x+y)=0 $$ $$ x(0)=1,\, x'(0)=0, \, x''(0)=5, \, y(0)=0, \, y'(0)=5 $$
I want to prove that there exists a unique solution, the domain of which includes a neighborhood of $t=0$.
If we set $$x_1=x,\,y_1=y$$ $$x_2=x',\, y_2=y'$$ $$x_3=x''$$ we have $$x_1'=x_2$$ $$x_2'=x_3$$ $$x_3'=-\frac{1}{1+t^2}x_3+\sin t\, x_2+\frac{t}{x_1^2+y_1^2+1}$$ and $$y_1'=y_2$$ $$y_2'=-e^{-t}y_2-\cos (x_1+y_1)$$ equivalent to $$\vec{x}'=\vec{F}(t,x,y)$$ $$\vec{y}'=\vec{G}(t,x,y)$$ Is there some way to merge these two systems and use Picard's theorem to complete the proof?
There's no need to build two distinct equations. Just set: $$ \begin{align*} &x_1=x \\ &x_2=x' \\ &x_3=x'' \\ &x_4=y\\ &x_5=y'\\ \end{align*} $$ The system tranforms to: $$ \begin{align*} &x_1'=x_2 \\ &x_2'=x_3 \\ &x_3'=-\frac{1}{1+t^2}x_3+\sin t\, x_2+\frac{t}{x_1^2+x_4^2+1} \\ &x_4'=x_5 \\ &x_5'=-e^{-t}x_5-\cos (x_1+x_4) \\ \end{align*} $$ Or, written more compactly: $$ \vec{x}'=\vec{F}(t,\vec{x}) $$ Then, there exists an arbitrary neighborhood $(-\delta,\delta)$ of the origin where $F$ is bounded.
This implies that the original system has a unique solution for $t \in (-\delta,\delta)$.