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Answers without the blur. Sign up and see all textbooks for free! Q 3.7-2E

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 ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{xy}}{\mathbf{-}}{{\mathbf{y}}}^{{\mathbf{2}}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.

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

See the step by step solution

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

Here $\mathrm{y}\text{'}=\mathrm{xy}-{\mathrm{y}}^{2},\mathrm{y}\left(0\right)=-1$

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 $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{xy}-{\mathrm{y}}^{2}$

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

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

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

The recursive formula is

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

for order p = 2 then

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

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_{0}=-1$.

Hence, the solution is role="math" localid="1664315352749" ${{\mathbf{y}}}_{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{+}\mathbf{h}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{{\mathbf{y}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}_{\mathbf{n}}}^{{\mathbf{2}}}\mathbf{\right)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}{\mathbf{\left(}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{+}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{-}}{\mathbf{2}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{\right)}\mathbf{\left(}{{\mathbf{x}}}_{{\mathbf{n}}}{{\mathbf{y}}}_{{\mathbf{n}}}{\mathbf{-}}{{\mathbf{y}}_{\mathbf{n}}}^{{\mathbf{2}}}\mathbf{\right)}\mathbf{\right)}$ ### Want to see more solutions like these? 