Chapter 9.3, Problem 35E

a.

To determine

To state: A formula for the function $y$ that does not involve any integral. The initial value problem is $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$ .

Answer to Problem 35E

The formula is not possible without using the integral.

Explanation of Solution

Given information:

It is given that the initial value problem is $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$ .

Consider the given initial value:

  $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$

It is not possible to find a formula for y without involving any integrals.

Therefore, the formula is not possible.

b.

To determine

To state: A power series that represent $y$ .

Answer to Problem 35E

The resultant answer is $T_n={\left(-1\right)}^{n-1}\frac{x^{2n-1}}{\left(2n-1\right)\left(n-1\right)}$ .

Explanation of Solution

Given information:

It is given that the initial value problem is $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$ .

Consider the initial value $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$ ,

Yes, the power series for y can be obtained. Power series generated by the function $e^x$ is given by:

  $e^x=1+x+\frac{x^2}{2!}+\cdots \cdots +\frac{x^n}{n!}+\cdots$

Replacing $x\text{ by }-x^2$ ,

  $\begin{array}{c}\frac{dy}{dx}=e^{-x^2}\\ =1+\left(-x^2\right)+\frac{{\left(-x^2\right)}^2}{2!}+\frac{{\left(-x^2\right)}^3}{3!}+\cdots +\frac{{\left(-x^2\right)}^n}{n!}+\cdots \\ =1-x^2+\frac{x^4}{2!}+\cdots +{\left(-1\right)}^n\frac{x^{2n}}{n!}+\cdots \end{array}$

The constant term of y is $y\left(0\right)=2$ , and the remaining terms of y can be obtained by integrating the above series.

  $y=2+x-\frac{x^3}{3}+\frac{x^5}{10}-\cdots +{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)n!}+\cdots$

By substituting $\left(n-1\right)$ for n , the general term can also be written as:

  $T_n={\left(-1\right)}^{n-1}\frac{x^{2n-1}}{\left(2n-1\right)\left(n-1\right)}$

c.

To determine

To state: The value of $x$ for which the power series actually equal the function y .

Answer to Problem 35E

The power series equals the function y for all real $x$ .

Explanation of Solution

Given information:

The given statement says to state the value of $x$ for which the power series actually equal the function y .

Consider the initial value $\frac{dy}{dx}=e^{-x^2}\text{ and }y=2\text{ when }x=0$ ,

Since the series for $e^{-x^2}$ converges for all real values for x .

Therefore, the series obtained by integrating also converges for all $x$ .

By theorem 2, thus the power series equals the function y for all real $x$ .