Series Solutions: Introduction

Exercise 1:

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


and thus a2=3/2.

Differentiating y''=3y yields


in particular


so a3=-3/3!=-1/2.

Since y(4)=3 y'', we get a4=9/4!=3/8.

One more time: y(5) = 3y''', so a5 = 3(- 3)/5! = - 3/40.

The Taylor approximation has the form

f (t) = 1 - t + $\displaystyle {\textstyle\frac{3}{2}}$t2 - $\displaystyle {\textstyle\frac{1}{2}}$t3 + $\displaystyle {\textstyle\frac{3}{8}}$t4 - $\displaystyle {\textstyle\frac{3}{40}}$t5.

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Helmut Knaust

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