Is this the Correct Equation for Approximating y(0.8) Using Euler's Method?

[alane1994]

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.

[TABLE="class: cms_table_grid, width: 500, align: center"]
[TR]
[TD="align: center"]

[FONT=MathJax_Main]Δ[FONT=MathJax_Math]t​

[/TD]
[TD="align: center"]Approximations of y(0.4)[/TD]
[TD="align: center"]Approximations of y(0.8)[/TD]
[/TR]
[TR]
[TD="align: center"]0.4[/TD]
[TD="align: center"]0.20000[/TD]
[TD="align: center"]?[/TD]
[/TR]
[TR]
[TD="align: center"]0.2[/TD]
[TD="align: center"][/TD]
[TD="align: center"][/TD]
[/TR]
[TR]
[TD="align: center"]0.1[/TD]
[TD="align: center"][/TD]
[TD="align: center"][/TD]
[/TR]
[TR]
[TD="align: center"]0.05[/TD]
[TD="align: center"][/TD]
[TD][/TD]
[/TR]
[/TABLE]

 

[MarkFL]

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

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

 

[MarkFL]

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]

I GOT IT! (Party)(Party)(Dance)(Dance)(Bigsmile)...(Cool)

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.