Question

In problems $\mathbf{1}\mathbf{-}\mathbf{4}$Use Euler’s method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{y}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{y}\mathbf{)}$,$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

$x_n$	0.1	0.2	0.3	0.4	0.5
$y_n$	2.7	2.511	2.383	2.291	2.225

Step-by-step solution

Writing the recursive formula

We have, $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}y\left(2-y\right)\mathbf{,}{\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$

Then,$$\begin{gathered} {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{h}·\mathbf{f}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\mathbf{)} \\ \mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{n}}{\mathbf{-}\mathbf{y}}_{\mathbf{n}}^{\mathbf{2}}\mathbf{)} \end{gathered}$$

Putting  n=0 to find  y1

$$\begin{gathered} {\mathbf{y}}_{\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{0}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{1}}{\mathbf{-}\mathbf{y}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{3}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{6}\mathbf{-}\mathbf{9}\mathbf{)} \\ \mathbf{=}\mathbf{3}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7} \end{gathered}$$

The value of ${\mathbf{y}}_{\mathbf{1}}=2.7$ for ${\mathbf{x}}_1\mathbf{=}\mathbf{0}.1$

Putting  n=1 to find  y2

$$\begin{gathered} {\mathbf{y}}_{\mathbf{2}}\mathbf{=}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{1}}{\mathbf{-}\mathbf{y}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{29}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{189}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{511} \end{gathered}$$

The value of ${\mathbf{y}}_2=2.511$ for ${\mathbf{x}}_2\mathbf{=}\mathbf{0}.2$

Putting  n=2 to find  y3

$$\begin{gathered} {\mathbf{y}}_{\mathbf{3}}\mathbf{=}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{2}}{\mathbf{-}\mathbf{y}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{511}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{5}\mathbf{.}\mathbf{022}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{305}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{511}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{128}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{383} \end{gathered}$$

The value of ${\mathbf{y}}_3=2.383$ for ${\mathbf{x}}_3\mathbf{=}\mathbf{0}.3$

Putting  n=3 to find  y4

$$\begin{gathered} {\mathbf{y}}_{\mathbf{4}}\mathbf{=}{\mathbf{y}}_{\mathbf{3}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{3}}{\mathbf{-}\mathbf{y}}_{\mathbf{3}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{383}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{4}\mathbf{.}\mathbf{766}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{679}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{383}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{091}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{291} \end{gathered}$$

The value of ${\mathbf{y}}_4=2.291$ for ${\mathbf{x}}_4\mathbf{=}\mathbf{0}.4$

Putting  n=4 to find y5

$$\begin{gathered} {\mathbf{y}}_{\mathbf{5}}\mathbf{=}{\mathbf{y}}_{\mathbf{4}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}{\mathbf{y}}_{\mathbf{4}}{\mathbf{-}\mathbf{y}}_{\mathbf{4}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{291}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{4}\mathbf{.}\mathbf{582}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{249}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{291}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{066}\mathbf{)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{225} \end{gathered}$$

The value of ${\mathbf{y}}_5=2.225$ for ${\mathbf{x}}_5\mathbf{=}\mathbf{0}.5$

Hence, the solution is

$x_n$	0.1	0.2	0.3	0.4	0.5
$y_n$	2.7	2.511	2.383	2.291	2.225