Why is f(y,t+h) Equal to y(t+2) in Implicit Euler Method?

[porcupineman23]

Hey
I don't understand this backward euler solution, in particular why the f(y,t+h) is equal to y(t+2)

 

[Chestermiller]

Hey
I don't understand this backward euler solution, in particular why the f(y,t+h) is equal to y(t+2)

Because it's wrong.

It should be:

y(t+2)=y(t)+y'(t+2)h

Since, in this problem y'=-y, you have:

y(t+2)=y(t)-y(t+2)h

Solving for y(t+2) gives:
$$y(t+2)=\frac{y(t)}{1+2h}$$
Chet

 

[D H]

Because it's wrong.

I agree with you there!

It should be:

y(t+2)=y(t)+y'(t+2)h

Actually, it should be y(t+h)=y(t)+h*y'(t+h). Given that y'(t)=-y(t), this becomes y(t+h)=y(t)-h*y(t+h), or y(t+h)=y(t)/(1+h).

Solving for y(t+2) gives:
$$y(t+2)=\frac{y(t)}{1+2h}$$
Chet

Better: $y(t+2)=\frac {y(t)}{3}$.

You've already set h=2.

 

[Chestermiller]

I agree with you there!

Actually, it should be y(t+h)=y(t)+h*y'(t+h). Given that y'(t)=-y(t), this becomes y(t+h)=y(t)-h*y(t+h), or y(t+h)=y(t)/(1+h).


Better: $y(t+2)=\frac {y(t)}{3}$.

You've already set h=2.

Thanks DH. I'm usually more careful about checking over what I've written before I submit my replies. I hope I didn't confuse the OP too much.

Chet