Series Solutions: Introduction

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

$\begin{displaymath}y''-3y=0,\quad y(0)=1,\ y'(0)=-1.\end{displaymath}$

Since y(0)=1, a₀=1. Similarly y'(0)=-1 implies that a₁=-1.

Since y''=3y, we obtain

y''(0)=3y(0)=3,

and thus a₂=3/2.

Differentiating y''=3y yields

y'''=3y',

in particular

y'''(0)=3y'(0)=-3,

so a₃=-3/3!=-1/2.

Since y⁽⁴⁾=3 y'', we get a₄=9/4!=3/8.

One more time: y⁽⁵⁾=3 y''', so a₅=3(-1/2)/5!=-1/80.

The Taylor approximation has the form

$\begin{displaymath}f(t)=1-t+\frac{3}{2} t^2-\frac{1}{2}t^3+\frac{3}{8}t^4-\frac{1}{80}t^5.\end{displaymath}$