How is the solution for y[n+1] obtained using Euler's forward method?

[ChickenChakuro]

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!

 

[ChickenChakuro]

Bah, it is simple plug-and-chug. Should have known! Thanks!

 

[HallsofIvy]

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