Series Solutions: Introduction

Find the fourth degree Taylor polynomial of the solution to the differential equation

$\begin{displaymath}y''+y^2=0,\quad y(0)=1,\ y'(0)=0.\end{displaymath}$

The initial conditions yield a₀=1 and a₁=0. Since y''=-y², we obtain that

$\begin{displaymath}a_2=\frac{y''(0)}{2}=\frac{-1}{2}.\end{displaymath}$

Differentiating y''=-y² yields

$\begin{displaymath}y'''=-2y\cdot y',\end{displaymath}$

and hence a₃=0.

When we differentiate

$\begin{displaymath}y'''=-2y\cdot y',\end{displaymath}$

we obtain

$\begin{displaymath}y^{(4)}=-2 (y')^2-2y\cdot y'',\end{displaymath}$

and consequently

$\begin{displaymath}a_4=\frac{-2(0)-2(1)(-1)}{4!}=\frac{1}{12}.\end{displaymath}$

The fourth degree Taylor approximation has the form

$\begin{displaymath}f(t)=1-\frac{1}{2} t^2+\frac{1}{12}t^4.\end{displaymath}$

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Helmut Knaust
1998-06-29