How to solve this system of ODE: $ u'= - \frac{2v}{t^2}$ and $v'=-u $?

$$v''= -u' \implies v''t^2-2v=0$$ $$ v''t^2+2tv'-2tv'-2v=0$$ $$(t^2v')'-2(tv)'=0$$ $$(t^2v'-2tv)'=0$$ Integrate.

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Edit1

$$t^2v'-2tv=c_1$$ $$\left ( \dfrac v {t^2} \right)'=\dfrac {c_1}{t^4}$$ Integrate again and you are done. $$\left ( \dfrac v {t^2} \right)=\dfrac {k_1}{t^3}+k_2$$ $$\implies v(t)=\dfrac {k_1}{t}+k_2t^2$$