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Here I have a system of differential equations:

$u_{0}''=-1$

$u_{0}u_{0}''+u_{1}''=-1$

$u_{2}''+u_{1}''u_{0}+u_{0}''u_{1}=-1$

$u_{3}''+u_{2}''u_{0}+u_{1}''u_{1}+u_{2}u_{0}''=-1$

$u_{4}''+u_{0}u_{3}''+u_{2}''u_{1}+u_{2}u_{1}''+u_{4}u_{0}''=-1$

My initial conditions are that $u_{1}$ to $u_{4}$ are all 0 at t=0 and $u_{0}'(0)=1$ and $u_{1}'$ to $u_{4}'$ at t=0 are all 0

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  • $\begingroup$ I tried plugging in the values from the start and got to maybe part of the third line and I'm not so sure what method in solving ODEs I need to proceed. $\endgroup$ – cambelot Apr 26 '15 at 20:17
  • $\begingroup$ I just noticed that you have both $u_4''$ and $u_4$ in the final equation. Is this intentional? $\endgroup$ – jameselmore Apr 27 '15 at 1:58
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Hint:
$u_0'' = -1 \implies u_0 = -\frac12x^2 + Bx + C$
$u_0(0) = 0,\ u_0'(0) = 1 \implies u_0 = x - \frac12x^2$

$u_0u_0'' + u_1'' = -1 \implies u_1'' = -\frac12x^2 + x - 1 \implies u_1 = -\frac1{24}x^4 + \frac16x^3 - \frac12x^2 + Dx + E$
$u_1(0)=u'_1(0)=0\implies u_1 = \frac1{24}x^4 + \frac16x^3 - \frac12x^2$

If you continue in this way you should find that $u_k$ will be a polynomial of order $2^{k+1}$. Further, every step after the derivation of $u_1$ will introduce two new constants that will immediately be removed since $u_i(0) = u_i'(0)$, $i>0$.

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