Question

Determine the recursive formulas for the Taylor method of order 2 for the initial value problem$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{xy}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.

Short answer

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

Step-by-step solution

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.

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

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

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

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

Apply the recursive formulas for order 2

The recursive formula is

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

for order p = 2 then

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

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=-1$.

Hence, the solution is role="math" localid="1664315352749" ${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}{{\mathbf{y}}_{\mathbf{n}}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{(}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{-}\mathbf{2}{\mathbf{y}}_{\mathbf{n}}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{n}}{\mathbf{y}}_{\mathbf{n}}\mathbf{-}{{\mathbf{y}}_{\mathbf{n}}}^{\mathbf{2}}\mathbf{\left)}\mathbf{\right)}$