Question

Use the improved Euler’s method with tolerance to approximate the solution to $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at x= 1. For a tolerance of$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{003}$ , use a stopping procedure based on the absolute error.

Short answer

$\varphi \mathbf{\left(}1\mathbf{\right)}=0.716984$

Step-by-step solution

Find the equation of approximation value.

Here,$\mathrm{y}\text{'}=1-\mathrm{y}+{\mathrm{y}}^3,\mathrm{y}\left(0\right)=0$

For $\xi =0.01$ , x=0, y₀ = 0 , $\mathrm{c}=\pi$, M = 10, h = 1

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y}) \\ =1-\mathrm{y}+{\mathrm{y}}^3 \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =1-(\mathrm{y}+\mathrm{hF})+(\mathrm{y}+\mathrm{hF}{)}^3 \\ =1-\sin (\mathrm{y}+\mathrm{h}(1-\sin \mathrm{y}\left)\right) \\ 1-(\mathrm{y}+\mathrm{h}(1-\mathrm{y}+{\mathrm{y}}^3\left)\right)+(\mathrm{y}+\mathrm{h}(1-\mathrm{y}+{\mathrm{y}}^3){)}^3 \end{gathered}$$

Solve for x and y.

Apply initial points $\mathrm{x}=0,\mathrm{y}=0,\mathrm{h}=1$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=(\mathrm{x}+\mathrm{h}) \\ \mathrm{y}=\mathrm{x}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \end{gathered}$$

x = 1

y = 1

Hence $\varphi (1)=\mathrm{y}(1,1)=1$

Evaluate the value of x and y

$\mathrm{x}=0,\mathrm{y}=0,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=0.625 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0+0.5 \\ =0.5 \\ \mathrm{y}=0+0.25(1+0.625) \\ =0.40625 \end{gathered}$$

$\varphi (0.5)=0.40625$

Determine the value of x and t for the conditions.

$\mathrm{x}=0.5,\mathrm{y}=0.40625,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}=0.660798 \\ \mathrm{G}=0.663095 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0.5+0.5 \\ =1 \\ \mathrm{y}=0.40625+0.25(0.660798+0.663095) \\ =0.737223 \end{gathered}$$

$\varphi (1)=\mathrm{y}(1,0.5)=0.737223$

Determine the all-other values.

Apply the same procedure for all other values and the values are

$$\begin{gathered} \varphi (1)=\mathrm{y}(1,0.25) \\ =0.719412 \\ \varphi (1)=\mathrm{y}(1,0.125) \\ =0.716984 \end{gathered}$$

Since,

$$\begin{gathered} \left|\mathrm{y}(1,0.196350)-\mathrm{y}(1,0.392699)\right| \\ =\left|1.09580-1.09229\right| \\ =0.00351<0.01 \end{gathered}$$

$\varphi (1)=0.716984$

Hence the solution is role="math" localid="1664310510661" $\mathit{\varphi }\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{716984}$