Q4E - 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 $\mathbf{0}\mathbf{.}\mathbf{1}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$ .

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{x}}{\mathbf{y}}\mathbf{ }\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

$x_n$	0.1	0.2	0.3	0.4	0.5
$y_n$	-1	-1.01	-1.029	-1.085	-1.096

Write the recursive formula

For the solution use the Euler’s formula $y_{n+1}=y_n+h.f\left(x_n,y_n\right)$

Apply recursive formula

One has ,$f\left(x,y\right)=\frac{-x}{y}, x_0=0, y_0=-1, h=0.1$

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 y 1

Now, find the value of $y_1$ when n = 0, then

$$\begin{gathered} y_1=y_0+\left(0.1\right)\left(\frac{x_0}{y_0}\right) \\ =-1+\left(0.1\right)\left(0\right) \\ =-1 \end{gathered}$$

Hence, the value of  ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}$  when  ${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$

Put n = 1 to find y 2

 The value of $y_2$ is

$$\begin{gathered} y_2=y_1+\left(0.1\right)\left(\frac{x_1}{y_1}\right) \\ =-1+\left(0.1\right)\left(\frac{0.1}{-1}\right) \\ =-1.01 \end{gathered}$$

Thus, the value of  ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{01}$ when  ${\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$

Put n = 2 to find y 3

Now the value of  $y_3$ is

$$\begin{gathered} y_3=y_2+\left(0.1\right)\left(\frac{x_2}{y_2}\right) \\ =-1.01+\left(0.1\right)\left(\frac{0.2}{-1.01}\right) \\ =-1.01+\left(-0.019\right) \\ =-1.029 \end{gathered}$$

So, the value is  ${\mathit{y}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{029}$ when  ${\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$

Put n = 3 to find y 4

The value of $y_4$  is

$$\begin{gathered} y_4=y_3+\left(0.1\right)\left(\frac{x_3}{y_3}\right) \\ =-1.029+\left(0.1\right)\left(\frac{0.3}{-1.029}\right) \\ =-1.029+\left(-0.029\right) \\ =-1.058 \end{gathered}$$

Consequently, the value is  ${\mathit{y}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{058}$  when  ${\mathit{x}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$

Put n = 4 to find y 5

The value of $y_5$ is

$$\begin{gathered} y_5=y_4+\left(0.1\right)\left(\frac{x_4}{y_4}\right) \\ =-1.058+\left(0.1\right)\left(\frac{0.4}{-1.058}\right) \\ =-1.058+\left(-0.0378\right) \\ =-1.096 \end{gathered}$$

Therefore, the value is  ${\mathit{y}}_{\mathbf{5}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{096}$ when  ${\mathbf{x}}_{\mathbf{5}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$

Therefore the solution is

${\mathit{x}}_{\mathbf{n}}$	0.1	0.2	0.3	0.4	0.5
${\mathit{y}}_{\mathbf{n}}$	-1	-1.01	-1.029	-1.058	-1.096