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Q 3.7-1E

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 2 for the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{cos}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\right)}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\pi }}$.

${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hcos}\mathbf{\left(}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{\right)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbit{s}\mathbit{i}\mathbit{n}\left({x}_{n}+{y}_{n}\right)\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{\left(}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{\right)}\mathbf{\right)}$

See the step by step solution

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

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

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 $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right)$

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

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

## Step 2: Apply the recursive formulas for order 2

The recursive formula is

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

for order p = 2 then

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

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

Hence the solution is role="math" localid="1664314543589" ${{\mathbf{y}}}_{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{n}}}{\mathbf{+}}\mathbf{hcos}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{sin}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{\right)}$