Question

Use the improved Euler’s method with tolerance to approximate the solution to $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{tx}\mathbf{\right)}\mathbf{,}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$,at t= 1. For a tolerance of$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$ , use a stopping procedure based on the absolute error.

Short answer

$\varphi \text{(1) = 1.25495}$

Step-by-step solution

Find the equation of approximation value

Here $\frac{\mathrm{dx}}{\mathrm{dt}}=1+\mathrm{t}\sin \left(\mathrm{tx}\right),\mathrm{x}\left(0\right)=0$,

For $\xi =0.01$, x = 0, , c = 0, M = 10, h = 1

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{t},\mathrm{x}) \\ =1+\mathrm{tsin}\left(\mathrm{tx}\right) \\ \mathrm{G}=\mathrm{f}(\mathrm{t}+\mathrm{h},\mathrm{x}+\mathrm{hF}) \\ =1+(\mathrm{t}+\mathrm{h})\sin \left(\right(\mathrm{t}+\mathrm{h}\left)\right(\mathrm{x}+\mathrm{hF}\left)\right) \end{gathered}$$

Solve for x and t

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

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1.84147 \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_{\mathrm{n}+1}=({\mathrm{t}}_{\mathrm{n}}+\mathrm{h}) \\ {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \\ \mathrm{t}=1 \\ \mathrm{x}=1.42074 \end{gathered}$$

Hence $\varphi \left(1\right)=\mathrm{y}(1,1)=1.42074$

Evaluate the value of x and t

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

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1.1237 \end{gathered}$$

$$\begin{gathered} \mathrm{t}=0+0.5 \\ =0.5 \\ \mathrm{x}=0+0.25(1+1.1237) \\ =0530925 \\ \varphi (0.5)=0.530925 \end{gathered}$$

Determine the value of x and t for the conditions.

$\mathrm{t}=0.5,\mathrm{x}=0.530925,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}(0.5,0.530925)=1.13118 \\ \mathrm{G}(0.5,0.530925)=1.88962 \end{gathered}$$

$$\begin{gathered} \mathrm{t}=0.5+0.5 \\ =1 \\ \mathrm{x}=0.530925+0.25(1.13118+1.88962) \\ =1.28613 \\ \varphi (1)=\mathrm{y}(1,0.5)=1.28613 \end{gathered}$$

Determine the all-other values.

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

$$\begin{gathered} \varphi (1)=\mathrm{y}(1,0.25)=1.26026 \\ \varphi (1)=\mathrm{y}(1,0.125)=1.25495 \end{gathered}$$

Since,

$$\begin{gathered} \left|\mathrm{y}(1,0.125)-\mathrm{y}(1,0.25)\right|=\left|1.25495-1.26026\right| \\ =0.00531<0.01 \\ \varphi \left(\mathrm{t}\right)=1.25495 \end{gathered}$$

Hence the solution is role="math" localid="1664306405053" $\mathit{\varphi }\mathbf{\text{(1) = 1.25495}}$