Question

In Problems 5-10use a numerical solver and Euler’s method to obtain a four-decimal approximation of the indicated value. First use h=0.1and then use h=0.05.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{x}\mathit{y}\mathbf{+}\sqrt{\mathbf{y}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$

Short answer

The indicated value for h=0.1 is $\mathrm{y}(0.5)\approx 1.6902$, and for h=0.05 is $\mathrm{y}(0.5)\approx 1.7219$.

Step-by-step solution

Define Euler’s method.

The general form of the Euler’s method is given by,${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)$ where${\mathrm{x}}_{\mathrm{n}}={\mathrm{x}}_0+\mathrm{nh}$ and $\mathrm{n}=0,1,2,\dots$.

Find the indicated value for h=0.1.

Let the given be $\mathrm{h}=0.1,{\mathrm{x}}_0=0,$and${\mathrm{y}}_0=1$.

When n=0, then

$\begin{array}{c}{\mathrm{y}}_{0+1}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\\ {\mathrm{y}}_1={\mathrm{y}}_0+\mathrm{h}\left({\mathrm{x}}_0{\mathrm{y}}_0+\sqrt{{\mathrm{y}}_0}\right)\\ {\mathrm{y}}_1=1+0.1(0\times 1+\sqrt{1})\\ {\mathrm{y}}_1=1.1\end{array}$

When n=1, then

$\begin{array}{c}{x}_1=x_0+1\left(0.1\right)\\ =0+0.1\\ =0.1\end{array}$

Hence, by Euler’s method, it is given by

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2={\mathrm{y}}_1+\mathrm{h}\left({\mathrm{x}}_1{\mathrm{y}}_1+\sqrt{{\mathrm{y}}_1}\right)\\ {\mathrm{y}}_2=1.1+0.1(0.1\times 1.1+\sqrt{1.1})\\ {\mathrm{y}}_2=1.2159\end{array}$

Tabulate the indicated values for h=0.1.

Further continuing using the numerical solver and Euler’s method, the results are given in the table.

Find the indicated value for h-0.05.

Let the given be $\mathrm{h}=0.05,{\mathrm{x}}_0=0,$$y_0=1$and .

When n=0, then

$\begin{array}{c}{\mathrm{y}}_{0+1}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\\ {\mathrm{y}}_1={\mathrm{y}}_0+\mathrm{h}\left({\mathrm{x}}_0{\mathrm{y}}_0+\sqrt{{\mathrm{y}}_0}\right)\\ {\mathrm{y}}_1=1+0.05(0\times 1+\sqrt{1})\\ {\mathrm{y}}_1=1.05\end{array}$

When n=1, then

$\begin{array}{c}{\mathrm{x}}_1={\mathrm{x}}_0+1(0.05)\\ =0+0.05\\ =0.05\end{array}$

Hence, by Euler’s method, it is given by

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2={\mathrm{y}}_1+\mathrm{h}\left({\mathrm{x}}_1{\mathrm{y}}_1+\sqrt{{\mathrm{y}}_1}\right)\\ {\mathrm{y}}_2=1.05+0.05(0.05\times 1.05+\sqrt{1.05})\\ {\mathrm{y}}_2=1.1039\end{array}$

Tabulate the indicated values for h=0.05.

Further continuing using the numerical solver and Euler’s method, the results are given in the table.