Question

In Problems 5-10use a numerical solver and Euler’s method to obtain a four-decimal approximation of the indicated value. First use$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$and then use$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{y}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$

Short answer

The indicated value for h=0.1 is $\mathrm{y}(0.5)\approx 0.4198$, and for h=0.05 is $\mathrm{y}(0.5)\approx 0.4124$.

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 ${\mathrm{e}}^{-\mathrm{y}}$, 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

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

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

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2=0.1+(0.1){\mathrm{e}}^{-{\mathrm{y}}_1}\\ =0.1+(0.1){\mathrm{e}}^{-0.1}\\ \approx 0.1905\end{array}$

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

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

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

$\begin{array}{c}{\mathrm{y}}_{1+1}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ {\mathrm{y}}_2=0.05+(0.05){\mathrm{e}}^{-{\mathrm{y}}_1}\\ =0.05+(0.05){\mathrm{e}}^{-0.05}\\ \approx 0.0976\end{array}$

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.