Answer
(a) There are parts of the graph with a negative slope. Therefore, the function with the given graph can not be a solution of the differential equation.
(b) The slope at the point $(1,1)$ is positive. Therefore, the function with the given graph can not be a solution of the differential equation.
Work Step by Step
(a) $\frac{dy}{dt} = e^t(y-1)^2$
According to the equation, the slope $\frac{dy}{dt} \geq 0$ for all values of $t$ and $y$. However, we can see that there are parts of the graph with a negative slope. Therefore, the function with the given graph can not be a solution of the differential equation.
(b) $\frac{dy}{dt} = e^t(y-1)^2$
According to the equation, when $y=1$, then the slope $\frac{dy}{dt} = 0$. However, on the graph, we can see that the slope at the point $(1,1)$ is positive. Therefore, the function with the given graph can not be a solution of the differential equation.
