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Q 3.7-4E

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 ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ .

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

See the step by step solution

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

Here $\mathrm{y}\text{'}={\mathrm{x}}^{2}+\mathrm{y},\mathrm{y}\left(0\right)=0$

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 ${\text{f(x,y) = x}}^{\text{2}}\text{+ y}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\left(\mathrm{x},\mathrm{y}\right)=2\mathrm{x}\phantom{\rule{0ex}{0ex}}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)=1\phantom{\rule{0ex}{0ex}}$

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

## Step 2: Evaluate the values of f3(x,y)  and  f4 (x,y)

Apply the same procedure as step 1

${\mathrm{f}}_{3}\left(\mathrm{x},\mathrm{y}\right)=2+2\mathrm{x}+{\mathrm{x}}^{2}+\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{4}\left(\mathrm{x},\mathrm{y}\right)=2+2\mathrm{x}+\mathrm{x}+\mathrm{y}$

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

The recursive formula is

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

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

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

Hence the solution is

role="math" localid="1664317215716" ${{\mathbf{y}}}_{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{+}\mathbf{h}\mathbf{\left(}{{\mathbf{x}}_{\mathbf{n}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{+}{{\mathbf{x}}}^{{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{\left(}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}{{\mathbf{x}}}^{{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{\left(}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}{{\mathbf{x}}}^{{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{\right)}$