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.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$

Short answer

The indicated value for h=0.1 is $\mathrm{y}(0.5)\approx 1.8371$, and for h=0.05 is $\mathrm{y}(0.5)\approx 1.9332$.

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 $\mathrm{n}=0,1,2,\dots$. Substitute $\mathrm{f}(\mathrm{x},\mathrm{y})$by ${\mathrm{x}}^2+{\mathrm{y}}^2$, then the equation becomes, ${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+{\mathrm{he}}^{-{\mathrm{y}}_{\mathrm{n}}}$.

Find the indicated value for h=0.1

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

When n=0, then

$\begin{array}{c}{\mathrm{y}}_{0+1}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\\ {\mathrm{y}}_1=1+(0.1)\left({\mathrm{x}}_0^2{+\mathrm{y}}_0^2\right)\\ =1+(0.1)\left(0^2+1^2\right)\\ =1.1\end{array}$

When n=1, then

$\begin{array}{c}{\mathrm{x}}_1={\mathrm{x}}_0+1(0.1)\\ =0+0.1\\ =0.1\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2=1.1+(0.1)\left({\mathrm{x}}_1^2{+\mathrm{y}}_1^2\right)\\ =1.1+(0.1)\left(0.1^2+1.1^2\right)\\ =1.222\end{array}$

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 $\mathrm{h}=0.05,{\mathrm{x}}_0=0,$and ${\mathrm{y}}_0=0$.

When n=0, then

$\begin{array}{c}{\mathrm{y}}_{0+1}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\\ {\mathrm{y}}_1=1+(0.05)\left({\mathrm{x}}_0^2{+\mathrm{y}}_0^2\right)\\ =1+(0.05)\left(0^2+1^2\right)\\ =1.05\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2=1.05+(0.05)\left({\mathrm{x}}_1^2{+\mathrm{y}}_1^2\right)\\ =1.05+(0.05)\left(0.{05}^2+1.{05}^2\right)\\ \approx 1.1053\end{array}$

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.