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Euler forward equation

by ChickenChakuro
Tags: equation, euler, forward
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ChickenChakuro
#1
May15-09, 06:44 AM
P: 32
Hi all, I'm having trouble understanding a basic concept introduced in one of my lectures. It says that:

To solve the DE
[tex]y(t) + \frac{dy(t)}{dt} = 1[/tex] where [tex]y(t) = 0[/tex],

using the Euler (forward) method, we can approximate to:

[tex]y[n+1] = T + (1-T)y[n] [/tex] where [tex]T[/tex] is step size and [tex]y[0] = 0[/tex].

I have no idea how this result is obtained, the only thing they say is that in general for

[tex]\frac{dx_1}{dt} = \frac{x_1[n+1] - x_1[n]}{T}[/tex] for [tex]t = nT[/tex].

Can anyone please help me understand how they arrived at the solution for [tex]y[n+1][/tex]? Thanks!
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ChickenChakuro
#2
May15-09, 06:48 AM
P: 32
Bah, it is simple plug-and-chug. Should have known! Thanks!
HallsofIvy
#3
May15-09, 09:01 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,683
Four minutes! You didn't even give us a chance to explain!


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