Question

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

Short 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-by-step solution

Find the value of f2(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 ${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=-\sin (\mathrm{x}+\mathrm{y})(1+\cos (\mathrm{x}+\mathrm{y}\left)\right)$

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 isrole="math" localid="1664314543589" ${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hcos}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{sin}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{\left)}\mathbf{\right(}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{\left)}\mathbf{\right)}$