Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

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).

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

$x_n$	0.1	0.2	0.3	0.4	0.5
$y_n$	1.1	1.22	1.362	1.528	1.72

Writing the recursive formula

Given, $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}x+y\mathbf{,}{\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$

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

Putting n = 0 to find y 1

$$\begin{gathered} {\mathbf{y}}_{\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{0}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{0}}\mathbf{+}{\mathbf{y}}_{\mathbf{0}}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1} \end{gathered}$$

The value of ${\mathbf{y}}_{\mathbf{1}}=1.1$  for  ${\mathbf{x}}_1\mathbf{=}\mathbf{0}.1$

Putting n = 1 to find y 2

$$\begin{gathered} {\mathbf{y}}_{\mathbf{2}}\mathbf{=}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{y}}_{\mathbf{1}}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22} \end{gathered}$$

The value of  ${\mathbf{y}}_2=1.22$ for  ${\mathbf{x}}_2\mathbf{=}\mathbf{0}.2$

Putting n = 2 to find y 3

$$\begin{gathered} {\mathbf{y}}_{\mathbf{3}}\mathbf{=}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{y}}_{\mathbf{2}}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{142}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{362} \end{gathered}$$

The value of  ${\mathbf{y}}_3=1.362$ for  ${\mathbf{x}}_3\mathbf{=}\mathbf{0}.3$

Putting n = 3 to find y 5

$$\begin{gathered} {\mathbf{y}}_{\mathbf{4}}\mathbf{=}{\mathbf{y}}_{\mathbf{3}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}{\mathbf{y}}_{\mathbf{3}}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{362}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{362}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{362}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{166} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{528} \end{gathered}$$

The value of  ${\mathbf{y}}_4=1.528$ for  ${\mathbf{x}}_4\mathbf{=}\mathbf{0}.4$

Putting n = 4 to find y 5

$$\begin{gathered} {\mathbf{y}}_{\mathbf{5}}\mathbf{=}{\mathbf{y}}_{\mathbf{4}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}{\mathbf{x}}_{\mathbf{4}}\mathbf{+}{\mathbf{y}}_{\mathbf{4}}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{528}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{528}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{528}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{192} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{72} \end{gathered}$$

The value of  ${\mathbf{y}}_5=1.72$ for  ${\mathbf{x}}_5\mathbf{=}\mathbf{0}.5$

Hence, the solution is

${\mathit{x}}_{\mathbf{n}}$	0.1	0.2	0.3	0.4	0.5
${\mathit{y}}_{\mathbf{n}}$	1.1	1.22	1.362	1.528	1.72