- #1
ChickenChakuro
- 31
- 0
Hi all, I'm having trouble understanding a basic concept introduced in one of my lectures. It says that:
To solve the DE
y(t) + \frac{dy(t)}{dt} = 1 where y(t) = 0,
using the Euler (forward) method, we can approximate to:
y[n+1] = T + (1-T)y[n] where T is step size and y[0] = 0.
I have no idea how this result is obtained, the only thing they say is that in general for
\frac{dx_1}{dt} = \frac{x_1[n+1] - x_1[n]}{T} for t = nT.
Can anyone please help me understand how they arrived at the solution for y[n+1]? Thanks!
To solve the DE
y(t) + \frac{dy(t)}{dt} = 1 where y(t) = 0,
using the Euler (forward) method, we can approximate to:
y[n+1] = T + (1-T)y[n] where T is step size and y[0] = 0.
I have no idea how this result is obtained, the only thing they say is that in general for
\frac{dx_1}{dt} = \frac{x_1[n+1] - x_1[n]}{T} for t = nT.
Can anyone please help me understand how they arrived at the solution for y[n+1]? Thanks!
