Backward Euler Method for solving IVP

Hi, I have been trying to understand the Backward Euler Method for quite some time now so I have come to here for help. The question is as follows:

Compute y(1) with 2 significant digits of accuracy with y(x) defined by the initial value problem:

$\displaystyle y' + siny = x, y(0)=0$

I know I am supposed to be using backward Euler's method to find this (I can't use forward Euler method or Runge-Kutta why? because of the siny? because it's nonlinear?) I'm not trying to just get the answer (although that would be helpful) so you don't have to worry about that, but I would like an explanation if possible. So far I have, using a step size=1:

$\displaystyle y(1)=y(0) +f(1, y(1)) = Y $

$\displaystyle Y = 0 + (1-sinY)$

$\displaystyle F(Y) = 1-sinY-Y= 0$

$\displaystyle F'(Y)= -cosY-1$

$\displaystyle F''(Y)= sinY $

but I'm not sure how I am supposed to use this information to arrive at an answer.

any help would be greatly appreciated. Thanks!

Re: Backward Euler Method for solving IVP

Quote:

Originally Posted by

**trevman** Hi, I have been trying to understand the Backward Euler Method for quite some time now so I have come to here for help. The question is as follows:

Compute y(1) with 2 significant digits of accuracy with y(x) defined by the initial value problem:

$\displaystyle y' + siny = x, y(0)=0$

I know I am supposed to be using backward Euler's method to find this (I can't use forward Euler method or Runge-Kutta why? because of the siny? because it's nonlinear?) I'm not trying to just get the answer (although that would be helpful) so you don't have to worry about that, but I would like an explanation if possible. So far I have, using a step size=1:

$\displaystyle y(1)=y(0) +f(1, y(1)) = Y $

$\displaystyle Y = 0 + (1-sinY)$

$\displaystyle F(Y) = 1-sinY-Y= 0$

$\displaystyle F'(Y)= -cosY-1$

$\displaystyle F''(Y)= sinY $

but I'm not sure how I am supposed to use this information to arrive at an answer.

any help would be greatly appreciated. Thanks!

You have:

$\displaystyle y'=f(x,y)=x-\sin(y)$

Backward Euler gives:

$\displaystyle y(h)=y(0)+hf(h,y(h))$

so if we go for $\displaystyle y(1)$ in one step of backwards Euler we have:

$\displaystyle y(1)=y(0)+1\times f(1,y(1))=1-\sin(y(1))$

so $\displaystyle y(1)$ is the solution of $\displaystyle u=1-\sin(u)$

CB