MHB 242t.9.1.15 Use Euler's method to calculate the first three approximations

[karush]

$\tiny{242t.9.1.15}$
$\textsf{Use Euler's method to calculate the first three approximations}$
$\textsf{to the given initial value problem for the specified increment size.}$
$\textsf{Calculate the exact solution. Round to 4 decimal places.}$
\begin{align*}\displaystyle
\frac{dy}{dx}&=9x^8e^{-x^9}\, , y_{(0)}=8, \, dx=0.5 \\
\end{align*}
$\textit{$x_0=0$}$
\begin{align*}\displaystyle
y_1& =y_0 +\left(x_0 \cdot y_0 \right)\cdot dx \\
&=8+\left(-9(0)^8\cdot e^{0^9} \right)\cdot 0.5 =\color{red}{8}
\end{align*}
$\textit{ $x_1=x_0+dx$ then
$x_1 \therefore x_1=0.5$}$
\begin{align*}\displaystyle
y_2&=y_1+\left(9(x_1)^8 \cdot
e^{{(-x_1)}^8}
\right)\\
&=8+\left( 9(0.5)^8 \cdot e^{{-0.5}^9} \cdot 0.5 \right)=\color{red}{7.9825}
\end{align*}
$\textit{$x_2=x_{1}+0.5$ to then
$x_2 \therefore x_2=0.5+0,5=1$}$
\begin{align*}\displaystyle
y_3&=7.9825+\left( 9(1)^8 \cdot e^{{-1}^9} \cdot 0.5 \right)=\color{red}{6.327}
\end{align*}
\begin{align*}\displaystyle
y_{exact}&=\int 9x^8e^{-x^9} dx
=e^{-x^8}+C
=\color{red}{e^{-x^8}+7}
\end{align*}
$\textit{hopefully... suggestions}$

 

[MarkFL]

Your solution to the given IVP isn't correct...we are given:

$$\d{y}{x}=9x^8e^{-x^9}$$ where $$y(0)=8$$

$$\int_{y(0)}^{y(x)}\,du=\int_0^x 9v^8e^{-v^9}\,dv=-\int_0^{-x^9} e^w\,dw$$

$$y(x)=1+y(0)-e^{-x^9}$$

With $$y(0)=8$$, our solution is then:

$$y(x)=9-e^{-x^9}$$

Even without knowing the solution, we see the slope is positive for all $0<x$, so we should expect our approximations to be increasing as $x$ increases. Let's look at Euler's method for approximating $$y(1.5)=9-e^{-1.5^9}\approx9$$

$$y_0=8,\,x_0=0,\,\Delta x=0.5$$

$$y_1=8+0.5(9(0)^8e^{-0^9})=8$$

$$y_2=8+0.5\left(9(0.5)^8e^{-0.5^9}\right)\approx8.017543826230405$$

$$y_3=8.017543826230405+0.5\left(9(1)^8e^{-1^9}\right)\approx9.673001311501896$$

Can you explain why we got a value greater than 9 when we know:

$$\lim_{x\to\infty}y=9$$ ?

How could we improve this scheme for this function?