Chapter 8.RP, Problem 1RP

Find the first four nonzero terms in the Taylor polynomial approximation for the given initial value problem.

a. $y^{\prime }=xy-y^2$, $y\left(0\right)=1$

b. $z^{″}-x^3{z}^{\prime }+x{z}^2=0$; $z\left(0\right)=-1$, $z^{\prime }\left(0\right)=1$

To determine

(a)

To find:

The first four nonzero terms in the Taylor polynomial approximation for the given initial value problem.

Answer to Problem 1RP

Solution:

The first four terms in the Taylor approximation is,

$y\left(x\right)=1-x+\frac{3}{2}{x}^2-\frac{5}{3}{x}^3+\mathrm{...}$

Explanation of Solution

Given:

The initial value problem is,

$y^{\prime }=xy-y^2;y(0)=1$

Approach:

The formula for the Taylor polynomial of any degree centered at $x_0$, approximating a function $y\left(x\right)$ is given as,

$y\left(x\right)\approx y\left(x_0\right)+y^{\prime }\left(x_0\right)\left(x-x_0\right)+\frac{y^{″}(0)}{2!}{\left(x-x_0\right)}^2+\mathrm{...}$ …$(1)$

Calculation:

Consider the equation, $y^{\prime }=xy-y^2$ …$(2)$

Substitute $0$ for $x$ and $1$ for $y$ in equation $(2)$.

$\begin{array}{rll}{y}^{\prime }(0)& =& (0)(1)-1^2\\ & =& -1\end{array}$

Differentiate the equation $(2)$ with respect to $x$.

$y^{″}\left(x\right)=x{y}^{\prime }\left(x\right)+y\left(x\right)-2y\left(x\right)y^{\prime }\left(x\right)$ …$(3)$

Substitute $0$ for $x$, $1$ for $y(0)$ and $-1$ for $y^{\prime }(0)$ in equation $(3)$.

$\begin{array}{rll}{y}^{″}(0)& =& 0\cdot -1+1-2\cdot 1\cdot -1\\ & =& 3\end{array}$

Again differentiate equation $(3)$ with respect to $x$.

$y^{‴}\left(x\right)=x{y}^{″}\left(x\right)-2y\left(x\right)y^{″}\left(x\right)-2{\left(y^{\prime }\left(x\right)\right)}^2$ ... $(4)$

Substitute $0$ for $x$, $1$ for $y(0)$, $-1$ for $y^{\prime }(0)$, and $3$ for $y^{″}(0)$ in equation $(4)$.

$\begin{array}{rll}{y}^{‴}(0)& =& 0+2\cdot -1-2\cdot 1\cdot 3-2{\left(-1\right)}^2\\ & =& -8-2\\ & =& -10\end{array}$

Substitute $0$ for $x_0$, $1$ for $y(0)$, $-1$ for $y^{\prime }(0)$, $3$ for $y^{″}(0)$, and $-10$ for $y^{‴}(0)$ in equation $(1)$

$\begin{array}{lll}y\left(x\right)& \approx & y(0)+y^{\prime }(0)x+\frac{y^{″}(0)}{2!}{x}^2+\frac{y^{‴}(0)}{3!}{x}^3+\mathrm{...}\\ & =& 1-x+\frac{3}{2}{x}^2-\frac{10}{6}{x}^3+\mathrm{...}\\ & =& 1-x+\frac{3}{2}{x}^2-\frac{5}{3}{x}^3+\mathrm{...}\end{array}$

Therefore, the first four terms of the polynomial is,

$y\left(x\right)=1-x+\frac{3}{2}{x}^2-\frac{5}{3}{x}^3+\mathrm{...}$

Conclusion:

Hence, the first four terms in the Taylor approximation is,

$y\left(x\right)=1-x+\frac{3}{2}{x}^2-\frac{5}{3}{x}^3+\mathrm{...}$

To determine

(b)

To find:

The first four nonzero terms in the Taylor polynomial approximation for the given initial value problem.

Answer to Problem 1RP

Solution:

The first four terms in the Taylor approximation is,

$$

$z\left(x\right)=-1+x-\frac{x^3}{6}+\frac{x^4}{6}+\mathrm{...}$

Explanation of Solution

Given:

The initial value problem is,

$z^{″}-x^3{z}^{\prime }+x{z}^2=0;z(0)=-1,z^{\prime }(0)=1$

Calculation:

Consider the equation, $z^{″}-x^3{z}^{\prime }+x{z}^2=0$ …$(5)$

Substitute $-1$ for $z(0)$ and $1$ for $z^{\prime }(0)$ in equation $(5)$.

$\begin{array}{rll}{z}^{″}(0)-{\left(z(0)\right)}^3{z}^{\prime }(0)+0& =& 0\\ z^{″}(0)& =& {\left(-1\right)}^3\cdot 1\\ & =& 1\end{array}$

Differentiate the equation $(5)$ with respect to $x$.

$z^{‴}\left(x\right)-x^3{z}^{″}\left(x\right)-3x^2{z}^{\prime }+2xz{z}^{\prime }\left(x\right)+z^2=0$ …$(6)$

Substitute $0$ for $x$, $-1$ for $z(0)$, $1$ for $z^{\prime }(0)$, and $1$ for $z^{″}(0)$ in equation $(6)$.

$\begin{array}{rll}{z}^{‴}(0)-0\cdot z^{″}(0)-0+0+{\left(z(0)\right)}^2& =& 0\\ z^{‴}(0)+1& =& 0\\ z^{‴}(0)& =& -1\end{array}$

Again differentiate equation $(6)$ with respect to $x$.

$\left[\begin{array}{l}{z}^{(4)}\left(x\right)-6x{z}^{\prime }\left(x\right)-3x^2{z}^{‴}\left(x\right)-3x^2{z}^{″}\left(x\right)-x^3{z}^{‴}\left(x\right)+2xz\left(x\right)z^{\prime }\left(x\right)+2z\left(x\right)z^{\prime }\left(x\right)+2x{\left(z^{\prime }\right)}^2+2\left(z\left(x\right)\right)z^{\prime }\left(x\right)\end{array}\right]=0$ ... $(7)$

Substitute $0$ for $x$, $-1$ for $z(0)$, $1$ for $z^{\prime }(0)$, and $1$ for $z^{″}(0)$, and $-1$ for $z^{‴}(0)$ in equation $(7)$.

$\begin{array}{rll}{z}^{(4)}(0)-0+2\left(-1\right)(1)+2\left(-1\right)(1)& =& 0\\ z^{(4)}(0)-4& =& 0\\ z^{(4)}(0)& =& 4\end{array}$

Substitute $z$ for $y$, $0$ for $x_0$, $-1$ for $z(0)$, $1$ for $z^{\prime }(0)$, and $1$ for $z^{″}(0)$, $-1$ for $z^{‴}(0)$, and $4$ for $z^{(4)}(0)$ in equation $(1)$.

$\begin{array}{rll}z\left(x\right)& \approx & z(0)+z^{\prime }(0)x+\frac{z^{″}(0)}{2!}{x}^2+\frac{z^{‴}(0)}{3!}{x}^3+\frac{z^{(4)}(0)}{4!}+\mathrm{...}\\ & =& -1+x-\frac{x^3}{3!}+\frac{4}{4!}{x}^4+\mathrm{...}\\ & =& -1+x-\frac{x^3}{3!}-\frac{1}{6}{x}^4+\mathrm{...}\end{array}$

Therefore, the first four terms of the polynomial is,

$z\left(x\right)=-1+x-\frac{x^3}{6}+\frac{x^4}{6}+\mathrm{...}$

Conclusion:

Hence, the first four terms in the Taylor approximation is,

$$

$z\left(x\right)=-1+x-\frac{x^3}{6}+\frac{x^4}{6}+\mathrm{...}$