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Euler forward equation 
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#1
May1509, 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 + (1T)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! 


#2
May1509, 06:48 AM

P: 32

Bah, it is simple plugandchug. Should have known! Thanks!



#3
May1509, 09:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,339

Four minutes! You didn't even give us a chance to explain!



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