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.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{y}\mathbf{-}{\mathit{y}}^{\mathbf{2}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{;}\mathit{y}\left(0.5\right)$

Short answer

The indicated value for $h=0.1isy\left(0.5\right)\approx 0.6231,andforh=0.05isy\left(0.5\right)\approx 0.6228$.

Step-by-step solution

Define Euler’s method.

The general form of the Euler’s method is given by, ${\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}$where ${\mathit{x}}_{\mathbf{n}}\mathbf{=}{\mathit{x}}_{\mathbf{0}}\mathbf{+}\mathit{n}\mathit{h}$and $\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$.

Find the indicated value for h = 0.1.

Let the given be h = 0.1, and y₀ = 0.5.

When n = 0, then

$$\begin{gathered} y_{0+1}=y_0+hf\left(x_0,y_0\right) \\ {y}_1=y_0+h\left(y_0-y_0^2\right) \\ {y}_1=0.5+0.1\left(0.5-0.5^2\right) \\ {y}_1=0.5250 \end{gathered}$$

When n = 1, then

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

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

$$\begin{gathered} y_{1+1}=y_1+hf\left(x_1,y_1\right) \\ {y}_2=y_1+h\left(y_1-y_1^2\right) \\ {y}_2=0.5250+0.1\left(0.5250-0.{5250}^2\right) \\ {y}_2=0.5499 \end{gathered}$$

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.

xn	yn
0	0.5000
0.1	0.5250
0.2	0.5499
0.3	0.5747
0.4	0.5991
0.5	0.6231

Find the indicated value for h = 0.05.

Let the given be $h=0.05,x_0=0,and{y}_0=1$.

When n = 0, then

$$\begin{gathered} y_{0+1}=y_0+hf\left(x_0,y_0\right) \\ {y}_1=y_0+h\left(y_0-y_0^2\right) \\ {y}_1=0.5+0.05\left(0.5-0.5^2\right) \\ {y}_1=0.5125 \end{gathered}$$

When n = 1, then

$\begin{array}{rcl}{x}_1& =& x_0+1\left(0.05\right)\\ & =& 0+0.05\\ & =& 0.05\end{array}$

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

$$\begin{gathered} y_{1+1}=y_1+hf\left(x_1,y_1\right) \\ {y}_2=y_1+h\left(y_1-y_1^2\right) \\ {y}_2=0.5125+0.05\left(0.5250-0.{5250}^2\right) \\ {y}_2=0.5250 \end{gathered}$$

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.

xn	yn
0	0.5000
0.05	0.5125
0.1	0.5250
0.15	0.5375
0.2	0.5499
0.25	0.5623
0.3	0.5746
0.35	0.5868
0.4	0.5989
0.45	0.6109
0.5	0.6228