Determine the constants $a$ and $b$ in terms of the parameter $α$ in such a way that the multistep method have the largest order possible Determine the constants $a$ and $b$ in terms of the parameter $α$ in such a way that the multistep method 
$y_{n+1}=-ay_n - αy_{n-1} + hby'_n$
have the largest order possible. For what ranges of values ​​of $α$ is the stable method?
I am a little confused solving this problem, I do not know if what I have done is fine, I hope someone will tell me if I am going in the right direction, thank you very much.
To find the parameters in such a way that the multistep method has the largest possible order, do I have to pay attention to the global truncation error? A multistep method has the general form
$$y_{n+1}=\sum_{j=0}^{p}a_jy_{n-j}+h\sum_{j=-1}^{p}b_jf(x_{n-j},y_{n-j})$$ for all $n\geq p$.
And we know that for $\tau(h)=O(h^m)$ it is necessary and sufficient that 
$$\sum_{j=0}^{p}(-j)^ia_j+i\sum_{j=-1}^{p}(-j)^{i-1}b_j=1$$
For all $i=1,2,...,m$
What else could I do? I think that from here I can get conditions for $a$ and $b$ but I do not know what the $a_j$ and $b_j$ are. Here $p$ is the step size that I think is $2$.
For the second question I know that if the multistep method meets the root condition then it is stable. To see if it fulfills the root condition I have to take out the associated polynomial that is $\rho(r)=r^{p+1}-\sum_{j=0}^{p}a_jr^{p-j}$ and its roots. We know that in this case $p=2, a_0=-a$ and $a_1=-\alpha$, with which we have left that $\rho(r)=r^3+ar^2+\alpha r$ and so $r_0=0, r_1=\frac{-a+\sqrt{a^2-4\alpha}}{2}, r_2=\frac{-a-\sqrt{a^2-4\alpha}}{2}$ are the roots.
What I have to prove is that $|r_j|\leq 1$ for all $j=0,1,2$ and that if $|r_j|=1$ is for that $\rho'(r_j)\neq 0$ but I do not know how to do it, could someone help me please?
 A: 1) For the first part of the question, you are correct that you need to look at the global truncation error $\tau(h)$. Since you already know the necessary and sufficient condition for $\tau(h) = \mathcal{O}(h^m)$
$$ \sum_{j=0}^p (-j)^ia_j + i\sum_{j=-1}^p (-j)^{i-1}b_j = 1 \ \ \textrm{ for any $i=1,2,\dots,m,$} \tag{GTC}$$
you just need to find the coefficients $a_j, b_j$, substitute these into (GTC) and solve linear equations in terms of these coefficients. 
It should be clear that $p=1$. To see the required coefficients, let us expand the general form of multistep method with $p=1$ and compare it to the given multistep method:
\begin{align*}
y_{n+1} & = \sum_{j=0}^p a_jy_{n-j} + h\sum_{j=-1}^p b_jf(x_{n-j},y_{n-j}) \\
& = \sum_{j=0}^p a_jy_{n-j} + h\sum_{j=-1}^p b_jy_{n-j}' \\
& = a_0y_n + a_1y_{n-1} + h\left(b_{-1}y_{n+1}' + b_0y_n' + b_1y_{n-1}'\right) \\
& = -ay_n -\alpha y_{n-1} + hby_n'.
\end{align*}
We see that
\begin{align*}
a_0 & = -a \\
a_1 & = -\alpha \\
b_{-1} & = 0 \\
b_0 & = b \\
b_1 & = 0. 
\end{align*}
In particular, we need three equations since we only have three unknowns $a,\alpha,b$. Two equations comes from requiring the given method to be consistent. 

Theorem: The multistep method is consistent if and only if $\displaystyle\sum_{j=0}^1 a_j = 1$ and equation (GTC) holds with $i=1$, i.e. 
  $$ \sum_{j=0}^1 a_j = 1 \ \ \textrm{ and } \ \ -\sum_{j=0}^1 ja_j + \sum_{j=-1}^1 b_j = 1. $$
  I am using $p=1$ here again. 

The last equation comes from (GTC) with $i=2$. 
2) You are correct again, that you need to look at the root condition. You are almost correct except that $p$ should be 1 instead of 2. So the associated polynomial is 
\begin{align*}
\rho(r) & = r^{p+1} - \sum_{j=0}^p a_jr^{p-j} \\
& = r^2 - \sum_{j=0}^1 a_jr^{1-j} \\
& = r^2 + ar + \alpha
\end{align*}
which has roots 
$$ r = \frac{-a\pm\sqrt{a^2 - 4\alpha}}{2}. $$
Clearly this depends on both $a$ and $\alpha$ but since the question only asks for the range of $\alpha$, this means that $a$ is related to $\alpha$ in some way. Indeed, from the first part, one of the consistency conditions gives 
$$ \sum_{j=0}^1 a_j = a_0 + a_1 = -a - \alpha = 1 \implies a = -1-\alpha. $$
Substituting this into the roots $r$ then yields
\begin{align*}
r & = \frac{(1 + \alpha)\pm \sqrt{(1+\alpha)^2 - 4\alpha}}{2} \\
& = \frac{(1+\alpha)\pm \sqrt{1 + 2\alpha + \alpha^2 - 4\alpha}}{2} \\
& = \frac{(1+\alpha)\pm \sqrt{1 - 2\alpha + \alpha^2}}{2} \\
& = \frac{(1+\alpha)\pm \sqrt{(1-\alpha)^2}}{2} \\
& = \frac{(1+\alpha)\pm (1-\alpha)}{2}.
\end{align*}
The roots are 
$$ r_+ = \frac{1+\alpha+1-\alpha}{2} = 1 \ \ \textrm{ and } \ \ r_- = \frac{1+\alpha-(1-\alpha)}{2} = \alpha. $$
The final step is to check this against the root conditions and these should spit out ranges for $\alpha$, and I will let you finish this. 
