Euler's Method Continued

alane1994

Active member
OK, my homework system says that the 0.20000 is correct... but all of my others are incorrect...

Is this the correct term for the top right box?

$$y(0.8)=y(0.4)+(\Delta{t} \times y^{\prime}(0.4))$$
OR
$$y(0.8)=y(0.4)-(\Delta{t} \times y^{\prime}(0.4))$$

According to the 0.20000 answer... the top one is correct.

 [FONT=MathJax_Main]Δ[/FONT][FONT=MathJax_Math]t[/FONT]​ Approximations of y(0.4) Approximations of y(0.8) 0.4 0.20000 ????? 0.2 0.1 0.05

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MarkFL

Staff member
Since you have been given another problem, I'll work the original in full.

We are given:

$\displaystyle \frac{dy}{dx}=-2y$ where $\displaystyle y(0)=1$

I would generalize as follows:

Using Euler's method for this IVP, we find the recursion:

$\displaystyle y_{n+1}=y_{n}\left(1-2\Delta t \right)$ where $\displaystyle y_0=1$

which gives us the closed-form:

$\displaystyle y_n=\left(1-2\Delta t \right)^n$

a) To approximate $\displaystyle y(0.4)$, we find with:

i) $\displaystyle \Delta t=0.4\,\therefore\,n=1$ we have

$\displaystyle y(0.4)\approx y_1=1-2(0.4)=0.2$

ii) $\displaystyle \Delta t=0.2\,\therefore\,n=2$ we have

$\displaystyle y(0.4)\approx y_2=(1-2(0.2))^2=0.36$

iii) $\displaystyle \Delta t=0.1\,\therefore\,n=4$ we have

$\displaystyle y(0.4)\approx y_4=(1-2(0.1))^4=0.4096$

iv) $\displaystyle \Delta t=0.05\,\therefore\,n=8$ we have

$\displaystyle y(0.4)\approx y_8=(1-2(0.05))^8=0.43046721$

b) To approximate $\displaystyle y(0.8)$, we find with:

i) $\displaystyle \Delta t=0.4\,\therefore\,n=2$ we have

$\displaystyle y(0.4)\approx y_2=(1-2(0.4))^2=0.04$

ii) $\displaystyle \Delta t=0.2\,\therefore\,n=4$ we have

$\displaystyle y(0.4)\approx y_4=(1-2(0.2))^4=0.1296$

iii) $\displaystyle \Delta t=0.1\,\therefore\,n=8$ we have

$\displaystyle y(0.4)\approx y_8=(1-2(0.1))^8=0.16777216$

iv) $\displaystyle \Delta t=0.05\,\therefore\,n=16$ we have

$\displaystyle y(0.4)\approx y_{16}=(1-2(0.05))^{16}=0.185302018885$

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MarkFL

Staff member
Suppose we are given the IVP:

$\displaystyle \frac{dy}{dx}=ky$ where $\displaystyle 0\ne k\in{R}$ and $\displaystyle y(x_0)=y_0$

We then obtain using Euler's method:

$\displaystyle y_{n+1}=y_0(1+k\Delta x)^n$

where $\displaystyle \Delta x=\frac{x_n-x_0}{n}$

So, by solving the IVP, we should expect then that:

$\displaystyle y_0\lim_{n\to\infty}\left(1+\frac{k(x_n-x_0)}{n} \right)^n=y_0e^{k(x_n-x_0)}$

You should verify this is true using L'Hôpital's rule.

alane1994

Active member
I GOT IT!!!! .......

In the table at the top the $$y(0.4)=y(t_{k})$$.

You have the equation $$t_{k}=\Delta{t} \times k$$

You know the $$y(t_{k})$$ , as well as the $$\Delta{t}$$, you then just solve for "k"!

"k" is the number of iterations of it that you have to do.

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