How to solve this system of ODE: $ u'= - \frac{2v}{t^2}$ and $v'=-u $? I have this system of differential equations:
$$ \left\{\begin{array}{ccc}u'&=& - \dfrac{2v}{t^2}\,, \\ v'&=&-u
\,.\end{array}\right. $$
I want to find the general solution by deriving an Euler differential equation for $v$ and giving a fundamental system.
So $v''= -u' \implies v''t^2-2v=0 $.
I am having issues solving this Euler differential equation.  How do I proceed?
 A: $$v''= -u' \implies v''t^2-2v=0$$
$$ v''t^2+2tv'-2tv'-2v=0$$
$$(t^2v')'-2(tv)'=0$$
$$(t^2v'-2tv)'=0$$
Integrate.

Edit1
$$t^2v'-2tv=c_1$$
$$\left ( \dfrac v {t^2} \right)'=\dfrac {c_1}{t^4}$$
Integrate again and you are done.
$$\left ( \dfrac v {t^2} \right)=\dfrac {k_1}{t^3}+k_2$$
$$\implies v(t)=\dfrac {k_1}{t}+k_2t^2$$
A: Write $s:=\ln(t)$, and $V(s):=v\big(\exp(s)\big)$.  Therefore,
$$V'(s)=\exp(s)\,v'\big(\exp(s)\big)\,,$$
and
$$V''(s)=\frac{\text{d}}{\text{d}s}\,\Big(\exp(s)\,v'\big(\exp(s)\big)\Big)=\big(\exp(s)\big)^2\,v''\big(\exp(s)\big)+\exp(s)\,v'\big(\exp(s)\big)\,.$$
As $t^2\,v''(t)=2\,v(t)$, we get
$$V''(s)=2\,v\big(\exp(s)\big)+\exp(s)\,v'\big(\exp(s)\big)=2\,V(s)+V'(s)\,.$$
Therefore,
$$V''(s)-V'(s)-2\,V(s)=0\,.\tag{*}$$
You should be able to solve the differential equation above now.
Remark.  Due to an approach like the above, a typical strategy for a Cauchy-Euler differential equation such as the one in this question is to assume that there exists a constant $r$ such that $v(t)=t^r$ is a solution, as User8128 suggested.  After all, solutions $V(s)$ to (*) are in the span of solutions of the form $V(s)=\exp(rs)$ for some $r$.  Since $t=\exp(s)$, $$v(t)=V\big(\ln(t)\big)=\exp\big(r\ln(t)\big)=t^r$$ is a solution.
