Q 3.7-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the recursive formulas for the Taylor method of order 2 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\pi }$.

Step-by-Step Solution

Answer

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

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

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

 

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

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

 

 

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

Step 2: Apply the recursive formulas for order 2

The recursive formula is

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

for order p = 2 then

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

 

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

 

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