Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{4}$ , and $\mathbf{0}\mathbf{.}\mathbf{5}$ , using steps of size $\mathbf{0}\mathbf{.}\mathbf{1}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$ .

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{x}}{\mathbf{y}}$ , $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$

Step-by-Step Solution

Answer

$x_n$	0.1	0.2	0.3	0.4	0.5
$y_n$	4	3.998	3.992	3.985	3.975

Write the recursive formula

One has,  $f\left(x,y\right)=\frac{-x}{y}, x_0=0, y_0=4, h=0.1$ .

Then $$\begin{gathered} y_{n+1}=y_n+h.f\left(x_n,y_n\right) \\ =y_n+\left(0.1\right)\left(\frac{-x_n}{y_n}\right) \end{gathered}$$

Put n = 0 to find role="math" localid="1663928029230" y 1

Now, find the value of  $y_1$

$$\begin{gathered} y_2=y_1+\left(0.1\right)\left(\frac{-x_1}{y_1}\right) \\ =4+\left(0.1\right)\left(\frac{-0.1}{4}\right) \\ =4-0.0025 \\ =3.998 \end{gathered}$$

Thus, the value is  ${\mathbf{y}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{998}$  when  ${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$

Put n = 2 to find y 3

Now the value of  $y_3$

$$\begin{gathered} y_3=y_2+\left(0.1\right)\left(\frac{-x_2}{y_2}\right) \\ =3.9975+\left(0.1\right)\left(\frac{-0.2}{3.9975}\right) \\ =3.9975-0.005 \\ =3.992 \end{gathered}$$

So, the values get  ${\mathit{y}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{992}$ when  ${\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$

Put n = 3 to find y 4

The value of  $y_4$ is

$$\begin{gathered} y_4=y_3+(0.1)\left(\frac{-x_3}{y_3}\right) \\ =3.9925+(0.1)\left(\frac{-0.3}{3.9925}\right) \\ =3.9925-0.0075 \\ =3.985 \end{gathered}$$

Consequently, the value of  ${\mathbf{y}}_{\mathbf{4}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{985}$ when  ${\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$

Put n = 4 to find y 5

The value of  $y_5$ is

$$\begin{gathered} y_5=y_4+\left(0.1\right)\left(\frac{-x_4}{y_4}\right) \\ =3.9849+\left(0.1\right)\left(\frac{-0.4}{3.9849}\right) \\ =3.9849-0.0100 \\ =3.975 \end{gathered}$$

Therefore, the value of  ${\mathbf{y}}_{\mathbf{5}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{975}$ when ${\mathbf{x}}_{\mathbf{5}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$ .

Therefore, the solution is

${\mathit{x}}_{\mathbf{n}}$	0.1	0.3	0.4	0.5	0.6
${\mathit{y}}_{\mathbf{n}}$	4	3.998	3.992	3.985	3.975