Q 3.7-4E - 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 $\mathbf{y}\mathbf{\text{'}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ .

Step-by-Step Solution

Answer

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

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

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

 

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

 

$$\begin{gathered} \frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=2\mathrm{x} \\ \frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})=1 \end{gathered}$$

 

So, the equation is ${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=2\mathrm{x}+{\mathrm{x}}^2+\mathrm{y}$

Step 2: Evaluate the values of f 3 ( x , y )     and     f 4   ( x , y )

Apply the same procedure as step 1

 

$$\begin{gathered} {\mathrm{f}}_3(\mathrm{x},\mathrm{y})=2+2\mathrm{x}+{\mathrm{x}}^2+\mathrm{y} \\ {\mathrm{f}}_4(\mathrm{x},\mathrm{y})=2+2\mathrm{x}+\mathrm{x}+\mathrm{y} \end{gathered}$$

Step 3: Apply the recursive formulas for order 4

The recursive formula is

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

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

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

Hence the solution is

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