Question

Using Eulers reverse method, is the answer to y'= 3y+t^2, where y(1)=1, step size 1/2 equal to -9/4 or -17/4 for the actual value at 1/2?


I got -9/4, but someone in my class did it and got -17/4

What I did was...
y(n+1) = yn + h(f(n+1, yn+1))
y1=y0 +.5(3y+1 + (tn+1)^2)
y1=1 + .5(3y+1 + .5^2)

-.5y1=.5(.5^2)+1

=-9/4

Is my work correct, or did I do something wrong?

Is the answer in fact -9/4, or was the other guy correct?

Answer 1

eulters method is just to follow the slope along its merry way

Answer 2

Even when you have a derivative?

Answer 3

y' = slope at a given point

and you are told to start at the point 1,1

Xnew = x+step

Answer 4

Ynew = slope * Xnew + Yold

Answer 5

or is it y' times step?

Answer 6

Seems like you are doing the forward euler...

Answer 7

.... seems that way. what is a reverse euler? :)

Answer 8

the formula for it is

(yn+1 - yn)/(step) = f(tn+1, yn+1)

It gets hairy because you have yn+1 on both sides, and it is implicit, you have to solve for it yn+1

Answer 9

well, i cant say that rinds any bells for me :/

Answer 10

or rings even

Answer 11

I'm feeling the same way. I can't tell if I'm doing it right, and the other guy just got the wrong answer, or if it is the complete opposite.

here is the wiki on the backward euler

http://en.wikipedia.org/wiki/Backward_Euler_method

Answer 12

srry, i cant make that stuff out either :/

Answer 13

Forwards Euler is so easy. Backwards is weird!