6.4 Working with taylor series  (Page 4/11)

We look for a solution of the form

Differentiating this function term by term, we obtain

If y satisfies the equation $y\text{″}=xy,$ then

Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

More generally, for $n\ge 3,$ we have $n·\left(n-1\right)c_n=c_{n-3}.$ In fact, all coefficients can be written in terms of $c_0$ and $c_1.$ To see this, first note that $c_2=0.$ Then

For $c_5,c_6,c_7,$ we see that

Therefore, the series solution of the differential equation is given by

The initial condition $y\left(0\right)=a$ implies $c_0=a.$ Differentiating this series term by term and using the fact that $y^{\prime }\left(0\right)=b,$ we conclude that $c_1=b.$ Therefore, the solution of this initial-value problem is

1. The Maclaurin series for $e^{\text{−}{x}^2}$ is given by
   
   $\begin{array}{cc}\hfill e^{\text{−}{x}^2}& ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\text{−}{x}^2\right)}^n}{n\text{!}}}\hfill \\ & =1-x^2+\frac{x^4}{2\text{!}}-\frac{x^6}{3\text{!}}+\text{⋯}+{\left(\mathrm{-1}\right)}^n\frac{x^{2n}}{n\text{!}}+\text{⋯}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }{\left(\mathrm{-1}\right)}^n\frac{x^{2n}}{n\text{!}}.}\hfill \end{array}$
   
   Therefore,
   
   $\begin{array}{cc}\hfill {\displaystyle \int e^{\text{−}{x}^2}dx}& ={\displaystyle \int \left(1-x^2+\frac{x^4}{2\text{!}}-\frac{x^6}{3\text{!}}+\text{⋯}+{\left(\mathrm{-1}\right)}^n\frac{x^{2n}}{n\text{!}}+\text{⋯}\right)dx}\hfill \\ & =C+x-\frac{x^3}{3}+\frac{x^5}{5.2\text{!}}-\frac{x^7}{7.3\text{!}}+\text{⋯}+{\left(\mathrm{-1}\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)n\text{!}}+\text{⋯}.\hfill \end{array}$
2. Using the result from part a. we have
   
   ${\displaystyle {\int }_0^1{e}^{\text{−}{x}^2}dx}=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\text{⋯}.$
   
   The sum of the first four terms is approximately $0.74.$ By the alternating series test, this estimate is accurate to within an error of less than $\frac{1}{216}\approx 0.0046296<0.01.$