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Expert-verified Found in: Page 426 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # In the study of the vacuum tube, the following equation is encountered:${{\mathbit{y}}}^{{\mathbf{"}}}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{.}}{\mathbf{1}}{\mathbf{\right)}}\mathbf{\left(}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{\right)}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{+}}{\mathbit{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}$Find the Taylor polynomial of degree 4 approximating the solution with the initial values${\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$ , ${{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$.

The required polynomial is, ${p}_{3}\left(x\right)=1-\frac{{x}^{2}}{6}$.

See the step by step solution

## Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at ${x}_{0}$, approximating a function$f\left(x\right)$ possessing n derivatives at ,${x}_{0}$ is given by

${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\text{'}}\left({x}_{0}\right)×\left(x-{x}_{0}\right)+{f}^{\text{'}\text{'}}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+\cdots +{f}^{n}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{n}}{n!}$

It is given that for the function ,$y\left(x\right)$

$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

The Taylor's polynomial cantered around${x}_{0}=0$ is given by

${p}_{n}\left(x\right)=y\left(0\right)+{y}^{\text{'}}\left(0\right)×\left(x-0\right)+{y}^{\text{'}\text{'}}\left(0\right)×\frac{{\left(x-0\right)}^{2}}{2!}+\cdots +{y}^{n}\left(0\right)×\frac{{\left(x-0\right)}^{n}}{n!}$

We need the value of $y\left(0\right),{y}^{\text{'}}\left(0\right),{y}^{\text{'}\text{'}}\left(0\right)$and ${y}^{\text{'}\text{'}\text{'}}\left(0\right)$etc for finding the value of the four non zero terms. The first two are provided by the initial conditions.

The value of ${y}^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{c}{y}^{\text{'}\text{'}}=-y-\left(0.1\right)\left({y}^{2}-1\right){y}^{\text{'}}\\ {y}^{\text{'}\text{'}}\left(0\right)=-y\left(0\right)-\left(0.1\right)\left({y}^{2}\left(0\right)-1\right){y}^{\text{'}}\left(0\right)\\ =-1\end{array}$

Now since ${y}^{\text{'}\text{'}}=-y-\left(0.1\right)\left({y}^{2}-1\right){y}^{\text{'}}$holds for some interval around ${x}_{0}=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-{y}^{\text{'}}-\left(0.1\right)\left[\left({y}^{2}-1\right)\left({y}^{\text{'}\text{'}}\right)+2y{\left({y}^{\text{'}}\right)}^{2}\right]\\ {y}^{\text{'}\text{'}\text{'}}\left(0\right)=-{y}^{\text{'}}\left(0\right)-\left(0.1\right)\left[\left({\left(y\left(0\right)\right)}^{2}-1\right)\left({y}^{\text{'}\text{'}}\left(0\right)\right)+2y\left(0\right){\left({y}^{\text{'}}\left(0\right)\right)}^{2}\right]\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^{4}=-{y}^{\text{'}\text{'}}-\left(0.1\right)\left[\left({y}^{2}-1\right)\left({y}^{\text{'}\text{'}\text{'}}\right)+2y\left({y}^{\text{'}}\right)\left({y}^{\text{'}\text{'}}\right)+2{\left({y}^{\text{'}}\right)}^{3}+2y\left(2{y}^{\text{'}}\right)\left({y}^{\text{'}\text{'}}\right)\right]\\ {y}^{4}\left(0\right)=-{y}^{\text{'}\text{'}}\left(0\right)-\left(0.1\right)\left[\left(y{\left(0\right)}^{2}-1\right)\left({y}^{\text{'}\text{'}\text{'}}\left(0\right)\right)+2y\left(0\right)\left({y}^{\text{'}}\left(0\right)\right)\left({y}^{\text{'}\text{'}}\left(0\right)\right)+2{\left({y}^{\text{'}}\left(0\right)\right)}^{3}+2y\left(0\right)\left(2{y}^{\text{'}}\left(0\right)\right)\left({y}^{\text{'}\text{'}}\left(0\right)\right)\right]\\ =1\end{array}$

## Step 2:Final proof

Hence by substituting the Taylor polynomial of degree four for the solution is given by.${p}_{n}\left(x\right)=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}$ ### Want to see more solutions like these? 