I've been stuck with this problem for a couple of days:
Let $u(t),v(t),w(t)$ solutions to the differential equation $y'''+y=0$, such that $u(0)=1, u'(0)=u''(0)=0$, $v(0)=v''(0)=0, v'(0)=1$ and $w(0)=w'(0)=0, w''(0)=1$. Prove, without solving the equation, that i) $u'(t)=-w(t)$ ii) $v'(t)=w(t)$ iii) $w'(t)=v(t)$ iv) $W[u,v,w]=u^3-v^3+w^3+3uvw=1$ ($W$ is the Wronskian)
None of my tries to prove i) to iii) have been really succesful, since most of them involve solving the equation explicitly, in one way or another. I tried to extract information from the Wronskian since it relates the three solutions and their 1st and 2nd derivatives, however I didn't manage to obtain anything useful in that sense (just that it equals one...)
I'd appreciate any small hint to get started with this problem, but not the answer because I want to try myself.
For the frst question, if we have $u'(t)=-w(t)$ then we also have $u'(t)+w(t)=0$ and you can observe that zero is a solution of the original DE so: $$(u'(t)+w(t))'''+(u'(t)+w(t))=0$$ Since $w(t)$ is a solution we have $w'''(t)+w(t)=0$ we are left with: $$u''''(t)+u'(t)=0$$ Integrate: $$u'''(t)+u(t)=C$$ Since $u(t)$ is a solution of the DE we must have the constant $C=0$ $$u'''(0)+u(0)=0$$ We are given $u(0)=1$ so that we must have $u'''(0)=-1$ This is the case since: $$u'(t)=-w(t) \implies u'''(0)=-w''(0)=-1$$ Sine it's given that $w''(0)=1$.