Trace Determinant Plane Differential Eqns Consider the 2 parameter family of linear systems 
$$\frac{DY(t)}{Dt} = \begin{pmatrix}
                  a & 1 \\
                  b & 1  \end{pmatrix} Y(t)
$$
In the ab plane, identify all regions where this system posseses a saddle, a sink, a spiral sink, and so on. 
I was able to get the eigenvalues as $$\lambda = \frac{a+1}{2} \pm \frac{\sqrt{(a+1)^2 - 4(a-b)}}{2}$$
but need help in finding the sink and source.
I got the spiral sink as: if $a \lt -1$
spiral source if $a \gt -1$
and center if $a = -1$
Can someone check this?
 A: Summarizing the comments: the best way to begin is to look at determinant $a-b$ and trace $a+1$: 


*

*$a-b<0$: saddle

*$a-b> 0$ and $a+1=0$: stable center

*$a-b> 0$ and $a+1<0$: stable node or spiral, depending on $(a+1)^2-4(a-b)$ being positive or negative

*$a-b> 0$ and $a+1>0$: unstable node or spiral,  depending on $(a+1)^2-4(a-b)$ being positive or negative

A: Here is a slightly more high-brow answer. Given the trace-determinant plane (together with the qualitative behaviors associated to different regions of it), one can determine all cases in one unified argument.
Let us first be explicit on the dependencies of the coefficient matrix to the parameters and put
$$A_{a,b}=\begin{pmatrix} a&1\\b&1\end{pmatrix},\,\, a,b\in\mathbb{R}.$$
As noted above, $\operatorname{tr}(A_{a,b})=a+1$ and $\det(A_{a,b})=a-b$. Thus the point $(a+1,a-b)$ on the trace-determinant plane determines completely ($\ast$). Further, note that since there are no restrictions on $a$ and $b$ all of the trace-determinant plane is covered by varying $a$ and $b$. Since we're also asked about the $ab$-plane, this suggest that we define a coordinate change from the $ab$-plane to the $td$-plane (=trace-determinant plane). Define
$$\Phi:\mathbb{R}^2\to \mathbb{R}^2, \begin{pmatrix}a\\b\end{pmatrix}\mapsto \begin{pmatrix} a+1\\a-b\end{pmatrix}. $$
(It is structurally clear where $\Phi$ comes from. Indeed $\Phi(a,b)=(\operatorname{tr}(A_{a,b}),\det(A_{a,b}))$.)
Note that $\Phi \begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix} 1&0\\1&-1\end{pmatrix}\begin{pmatrix} a\\b\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}$, so that $\Phi$ is an affine coordinate change with inverse
$$\Phi^{-1}\begin{pmatrix}t\\d\end{pmatrix}=\begin{pmatrix} 1&0\\1&-1\end{pmatrix}\begin{pmatrix} t\\d\end{pmatrix}+\begin{pmatrix}-1\\-1\end{pmatrix}.$$
Here is a humble caricature for this coordinate change (https://www.desmos.com/calculator/afq5mwx6pc is the link for the $ab$-plane and the trace-determinant plane is from Hirsch, Smale and Devaney's Differential Equations, Dynamical Systems, and an Introduction to Chaos (3e, p.64)):

The behavioral regions on the trace-determinant plane are thus transferred to the $ab$-plane. All that's left is to calculate equations for the curves separating the regions, which is easily done by plugging in the equations for the corresponding curves in the trace-determinant plane. For instance the critical curve $d=\dfrac{t^2}{4}$ is transformed into $\Phi^{-1}\begin{pmatrix}t\\\frac{t^2}{4}\end{pmatrix}= \begin{pmatrix}t-1\\ t-1+\frac{t^2}{4}\end{pmatrix}$.

Regarding my statement above marked with ($\ast$), note that the trace determinant plane does not distinguish two distinct behaviors on the critical curve $\dfrac{t^2}{4}=d$, (if $\dfrac{t^2}{4}=d$, then the matrix $A_{a,b}=A_{t-1,t-1-d}=A_{t-1,t-1-\frac{t^2}{4}}$ may or may not be diagonalizable) hence for the complete analysis of the behaviors some further inquiry is required. Say $\dfrac{t^2}{4}=d$. Then the repeated eigenvalue of $A$ is $\dfrac{t}{2}$. The associated eigenvector equation is
$$\begin{pmatrix} 0\\0\end{pmatrix}=\begin{pmatrix} \frac{t}{2}-1 & 1\\ t-1-\frac{t^2}{4} & 1-\frac{t}{2}\end{pmatrix}\begin{pmatrix} v_1\\ v_2\end{pmatrix} =\begin{pmatrix} \frac{t}{2}-1 & 1\\ -\left(\frac{t}{2}-1\right)^2 & -\left(\frac{t}{2}-1\right)\end{pmatrix}\begin{pmatrix} v_1\\ v_2\end{pmatrix},$$
so that any eigenvector associated to $\dfrac{t}{2}$ is proportional to $V_t=\begin{pmatrix} 1\\ -\left(\frac{t}{2}-1\right)\end{pmatrix}$, whence
$$A_{t-1,t-1-\frac{t^2}{4}}\sim \begin{pmatrix} \frac{t}{2}&1\\0&\frac{t}{2}\end{pmatrix},$$
which completes the analysis.

Finally, here is a further exercise to exemplify the benefit of the qualitative method I've described here. Say the parameters $a$ and $b$ are chosen uniformly (and independently) at random from $[-2,2]$. Out of all phase portraits with positive likelihood, which phase portrait is the least likely? (Note that the coordinate change $\Phi$ is volume preserving.)
