Q 3.7-3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the recursive formulas for the Taylor method of order 4 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

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

Step 1: Find the value of f 2 (x,y)

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

 

Apply the chain rule.

 ${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})\mathrm{f}(\mathrm{x},\mathrm{y})$

 

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

 

$$\begin{gathered} \frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=1 \\ \frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})=-1 \end{gathered}$$

 

So, the equation is ${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=1-\mathrm{x}+\mathrm{y}$

Step 2: Evaluate the values of f 2 (x,y) and f 4 (x,y)

Apply the same procedure as step 1

 

$$\begin{gathered} {\mathrm{f}}_3(\mathrm{x},\mathrm{y})=-1+\mathrm{x}-\mathrm{y} \\ {\mathrm{f}}_4(\mathrm{x},\mathrm{y})=1-\mathrm{x}+\mathrm{y} \end{gathered}$$

Step 3: Apply the recursive formulas for order 4

The recursive formula is

$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+\frac{{\mathrm{h}}^2}{2!}{\mathrm{f}}_2({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+.....\frac{{\mathrm{h}}^{\mathrm{p}}}{\mathrm{p}!}{\mathrm{f}}_{\mathrm{p}}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}) \end{gathered}$$

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

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=0$.

Hence the solution is ${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{-}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}\mathbf{+}\mathbf{(}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}$