Answer
$\frac{dy}{dt}=y-\frac{1}{3}$, $a=-\frac{1}{3}$, $b=1$
Work Step by Step
For equations of the form $\frac{dy}{dt}=a+by$, all equilibrium solutions (called $y_{eq}$) can be found by solving the right side of the equation for $\frac{dy}{dt}=0$. The solution of this ($0=a+by$) is $y=-\frac{a}{b}$ (this is $y_{eq}$). If $b\lt0$, all values of $y$ such that $y{\ne}y_{eq}$ will be convergent. If $b\gt0$, all values of $y$ such that $y{\ne}y_{eq}$ will be divergent.
Our problem specifies that $y_{eq}=\frac{1}{3}$, therefore $-\frac{a}{b}=\frac{1}{3}$. Any values of $a$ and $b$ that satisfy this, and where $b\gt0$ will be solutions.
