Question

In Problem 20 proceed as in Example 3 and obtain the first six non zero terms of a Taylor series solution, centered at 0, of the given initial value problem. Use a numerical solver and a graphing utility to compare the solution curve with the graph of the Taylor polynomial.
$y\text{'}\text{'}=e^y,y\left(0\right)=1,y\text{'}\left(0\right)=-1$

Short answer

The solution of the above question is

$$\begin{gathered} y\left(x\right)=-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+ \\ \frac{1}{12}{x}^4-\frac{1}{24}{x}^5+\frac{1}{45}{x}^5 \end{gathered}$$

Step-by-step solution

Taylor’s theorem

A Taylor series is a way to approximate any function as a polynomial with an infinite number of terms. Each term of the Taylor polynomial comes from the function's derivatives at a single point.

$y\left(0\right)+\frac{y\text{'}\left(0\right)}{1!}x+\frac{y^n\left(0\right)}{2!}{x}^2+\frac{y^m\left(0\right)}{3!}{x}^3+\frac{y^4}{4!}{x}^4+\frac{y^5}{5!}{x}^5$

Taylor’s theorem will be applied;

We will apply the solution form of Taylor’s theorem in the following answer:

$$\begin{gathered} y\text{'}\text{'}=e^y \\ y\left(0\right)=1 \\ y\text{'}\left(0\right)=-1 \\ y\left(x\right)=y\left(0\right)+\frac{y\text{'}\left(0\right)}{1!}x+\frac{y\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{y\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{y^4\left(0\right)}{4!}{x}^4+\frac{y^5\left(0\right)}{5!}{x}^5+\frac{y^6\left(0\right)}{6}{x}^6 \\ y\text{'}\text{'}\left(0\right)=e^0 \\ =1 \\ y\text{'}\text{'}\text{'}\left(x\right)=e^{y}y\text{'} \\ y\text{'}\text{'}\text{'}\left(0\right)=e^0\times (-1) \\ =-1 \\ {y}^4\left(x\right)=e^y(y\text{'}{)}^2+e^{y}y\text{'}\text{'} \\ {y}^4\left(0\right)=1\times 1+1\times 1 \\ =2 \\ {y}^5\left(x\right)=e^y(y\text{'}{)}^3+2e^{y}y\text{'}y\text{'}\text{'}+e^{y}y\text{'}y\text{'}\text{'}+e^{y}y\text{'}\text{'}\text{'} \\ =e^y(y\text{'}{)}^3+3e^{y}y\text{'}y\text{'}\text{'}+e^{y}y\text{'}\text{'}\text{'} \\ {y}^5\left(0\right)=e^0(-1{)}^3+3e^0(-1)\left(1\right)+e^0(-1) \\ =-5 \end{gathered}$$

Finally, substituting with all the initial conditions, then we have:

$$\begin{gathered} y\left(x\right)=0+\frac{-1}{1}x+\frac{1}{2\times 1}{x}^2+\frac{-1}{3\times 2\times 1}{x}^3+\frac{2}{4\times 3\times 2\times 1}{x}^4+\frac{-5}{5\times 4\times 3\times 2\times 1}{x}^5+\frac{16}{6\times 5\times 4\times 3\times 2\times 1}{x}^6\backslash \mathrm{hfill} \\ =-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{12}{x}^4-\frac{1}{24}{x}^5+\frac{1}{45}{x}^5 \end{gathered}$$

Graph

Graph is included in the following question;

Hence, he final answer is$y\left(x\right)=-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{12}{x}^4-\frac{1}{24}{x}^5+\frac{1}{45}{x}^5$