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I should perhaps mention that its been awhile since i did any mathematics and this was given in class today and we were asked as practice to find more solutions to this IVP. Can anyone perhaps explain why v(t) and u(t) are solutions and how they were found?

IVP $x^{'} =3x^{(2/3)} $ $x(0) = 0$

The answers given are $u(t)=0$ and $v(t)=t^3$

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when $$u(t)=0 \to u'(t)=0 \to x=u(t) \to 0=3*(0)^{\frac{2}{3}}$$ and $$u(t)=t^3 \to u'(t)=3t^2 \to x=u(t) \to 3t^2=3*(t^3)^{\frac{2}{3}}$$

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    I literally have no idea why your looking at $u^{'}(t)$ – 2017-01-11
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    So basically im looking for a function that's derivative equals the above equation at the point x(0)=0 ? – 2017-01-11
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    Because you have derivitative of x=3*x^(2/3) – 2017-01-11
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    $x(0)=0 $ is trivial solution of your IVP (initial value problem ) – 2017-01-11
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    wouldnt any $t^{n} : n>1 $ and an integer work? – 2017-01-11