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Ho can we obtain the (approximate) solution for, a $2$nd order differential equation, using successive approximations?

All the info I found so far explains how to do it for a 1st order equation. Can you point me to the general method for a $2$nd order case?

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Transform the second order equation $y''=f(x,y,y')$ in a first order system, introducing a new unknown $z=y'$: $$\begin{align} y'&=z\\ z'&=f(x,y,z)\\ y(x_0)&=y_0\\ z(x_0)&=z_0 \end{align}$$ The successive iterations are defined by $$\begin{align} y_{n+1}&=y_0+\int_{x_0}^xz_n(t)\,dt\\ z_{n+1}&=z_0+\int_{x_0}^xf(t,y_n(t),z_n(t))\,dt \end{align}$$

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  • $\begingroup$ Thank you. A minor detail: what does it mean to expand r to second order as r = r0 + ωr1 + ω²r2? Do I have to simply solve for the r0, r1 and r2 and then introduce them in that expression? $\endgroup$ – carllacan Jan 29 '14 at 20:37

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