Chapter 7.1, Problem 11E

a.

To determine

To explain : Why the function with the given graph can’t be solution of the given differential equation.

Explanation of Solution

Given information :

Differential equation: $\frac{dy}{dt}=e^t{\left(y-1\right)}^2\mathrm{.............}\left(i\right)$

  

Concept used :

If ${\frac{dy}{dt}|}_{\text{at some point}}=0$ then, slope of graph will be flat at that point.

The slope of the graph will be increasing or decreasing according to $\frac{dy}{dt}>0\text{ or }<0$

Calculation :

  $\frac{dy}{dt}=e^t{\left(y-1\right)}^2\mathrm{.............}\left(i\right)$

In equation (i) ${\left(y-1\right)}^2$ cannot be negative and $e^t$ can never be negative.

i.e. $\frac{dy}{dt}$ can never be negative.

It implies that $\frac{dy}{dt}>0$ other than $y=1$

Therefore, the slope of the graph cannot be negative.

But, in the first graph we can see that at some point slope of the graph is negative. Therefore, the graph (i) cannot be the solution of the differential equation.

b.

To determine

To explain : Why the function with the given graph can’t be solution of the given differential equation.

Explanation of Solution

Given information :

Differential equation: $\frac{dy}{dt}=e^t{\left(y-1\right)}^2\mathrm{.............}\left(i\right)$

  

Concept used :

If ${\frac{dy}{dt}|}_{\text{at some point}}=0$ then, slope of graph will be flat at that point.

Calculation :

  $\frac{dy}{dt}=e^t{\left(y-1\right)}^2\mathrm{.............}\left(i\right)$

  ${\frac{dy}{dt}|}_{\text{at }y=1}=0$

i.e. at $y=1$ the graph should be flat i.e. the graph will be horizontal

But there is no flat portion at $y=1$ in the graph. Hence graph is not the solution of the given solution of the given differential equation.