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Answers without the blur. Sign up and see all textbooks for free! Q 3.7-3E

Expert-verified Found in: Page 139 ### 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 # Determine the recursive formulas for the Taylor method of order 4 for the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$.

${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\mathbf{\left(}{\mathbf{x}}_{\mathbf{n}}\mathbf{-}{\mathbf{y}}_{\mathbf{n}}\mathbf{\right)}\mathbf{+}\left(\frac{{h}^{2}}{2}-\frac{{h}^{3}}{6}+\frac{{h}^{4}}{24}\right)\mathbf{\left(}\mathbf{1}\mathbf{-}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{\right)}$

See the step by step solution

## Step 1: Find the value of f2(x,y)

Here $\mathrm{y}\text{'}=\mathrm{x}-\mathrm{y},\mathrm{y}\left(0\right)=0$

Apply the chain rule.

${\mathrm{f}}_{2}\left(\mathrm{x},\mathrm{y}\right)=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\left(\mathrm{x},\mathrm{y}\right)+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$

Since $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}-\mathrm{y}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\left(\mathrm{x},\mathrm{y}\right)=1\phantom{\rule{0ex}{0ex}}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)=-1$

So, the equation is ${\mathrm{f}}_{2}\left(\mathrm{x},\mathrm{y}\right)=1-\mathrm{x}+\mathrm{y}$

## Step 2: Evaluate the values of f2(x,y) and f4(x,y)

Apply the same procedure as step 1

${\mathrm{f}}_{3}\left(\mathrm{x},\mathrm{y}\right)=-1+\mathrm{x}-\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{4}\left(\mathrm{x},\mathrm{y}\right)=1-\mathrm{x}+\mathrm{y}$

## Step 3: Apply the recursive formulas for order 4

The recursive formula is

${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}\left({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}\right)+\frac{{\mathrm{h}}^{2}}{2!}{\mathrm{f}}_{2}\left({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}\right)+.....\frac{{\mathrm{h}}^{\mathrm{p}}}{\mathrm{p}!}{\mathrm{f}}_{\mathrm{p}}\left({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}\right)$

${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{h}\left({\mathrm{x}}_{\mathrm{n}}-{\mathrm{y}}_{\mathrm{n}}\right)+\left(\frac{{\mathrm{h}}^{2}}{2}-\frac{{\mathrm{h}}^{3}}{6}+\frac{{\mathrm{h}}^{4}}{24}\right)\left(1-{\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}\right)$

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_{0}=0$.

Hence the solution is role="math" localid="1664316175651" ${{\mathbf{y}}}_{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{+}\mathbf{h}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{-}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}{\mathbf{\right)}}$ ### Want to see more solutions like these? 