Series Solutions: Introduction

Exercise 2:

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}


Answer.

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

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