# Explicit formula for the implicit Euler method

Given the problem;

$\displaystyle\cases{ y'(t)=y^2(t) & \cr y(0)=1 }$ for $t\in[0,1]$

Using the implicit euler method, find an explicit formula to get $y_{n+1}$

HINT: The method has to be consistent.

The problem is a special case of Bernoulli equation with the solution $y(t)=\frac{1}{1-t}$, now do I have to find an approximate formula without recurrence relation, is that correct ?

Consistency is defined as: $\epsilon_n=y(t_{n+1})-y(t_n)-h_nf(t_{n+1},y(t_{n+1}))$ for the implicit euler method, but since I must get here an explicit formula, what does it mean ?

• My guess: You write up the formula for implicit Euler and try to solve it for $y_{n+1}$. You get a quadratic equation, so you have to pick one of its roots. That is where the consistency comes in. – Harald Hanche-Olsen Oct 22 '14 at 15:33
• @Harald Hanche-Olsen I don't know if I'm on the right track but, I got $y_{n+1}=y(t_0)+\sum_{i=0}^{n}h_iy^2(t_{i+1})$ – derivative Oct 22 '14 at 15:44
• I think whoever posed the question only wants you to write an explicit formula for $y_{n+1}$ in terms of $y_n$, not an explicit formula in terms of $n$ and the initial data, which is much harder. But if I'm right, this should have been stated more clearly. – Harald Hanche-Olsen Oct 22 '14 at 16:50
• @Harald Hanche-Olsen the assistant has only mentioned that, it has to do something with taylor formula – derivative Oct 22 '14 at 18:51

With steplength $h$, the implicit Euler method for this equation would say $$y_{n+1}-y_n=hy_{n+1}^2.$$ Seeing this as a quadratic equation to solve for $y_{n+1}$, we get $$y_{n+1}=\frac{1\pm\sqrt{1-4hy_n}}{2h},$$ but you need to pick the correct sign. For concistency, you absolutely need to have $y_{n+1}$ converging to $y_n$ if you let $h\to0$, and that immediately eliminates the plus sign (which would have the right hand side going off to $\pm\infty$), leaving $$y_{n+1}=\frac{1-\sqrt{1-4hy_n}}{2h}$$ as the only sensible choice.
You may, and probably should, go further and verify that this is in fact consistent. This might involve taking a Taylor expansion of the square root, or perhaps my favourite trick for fractions of this kind: Multiplying numerator and denominator by $1+\sqrt{1-4hy_n}$. I'll stop here, as you do need to do some if this work on your own.