$($a$)$ What can you say about a solution of the equation $y^{\text{'}}=-\frac{1}{4}{y}^2$ just by looking at the differential equation?

The function $y$ must be equal to $0$ on any interval on which it is defined.

The function $y$ must be strictly decreasing on any interval on which it is defined.

The function $y$ must be increasing $($or equal to $0)$ on any interval on which it is defined.

The function $y$ must be decreasing $($or equal to $0)$ on any interval on which it is defined.

The function $y$ must be strictly increasing on any interval on which it is defined.

$($b$)$ Verify that all members of the family $y=\frac{4}{(xC)}$ are solutions of the equation in part $($a$).$

We substitute the values of $y$ and $y^{\text{'}}$ and test the solution to see if the left hand side $($UHS$)$ is equal to the right hand side $($RHS$).$

$y=\frac{4}{xC}=>y^{\text{'}}=-\frac{?}{(xC{)}^2}$

$LHS=y^{\text{'}}=-\frac{?}{(xC{)}^2}=-\frac{1}{4}(\frac{?}{xC}{)}^2=-\frac{1}{4}{y}^2=RHS$

$($c$)$ Can you think of a solution of the differential equation $y^{\text{'}}=-\frac{1}{4}{y}^2$ that is not a member of the family in part $($b$)?$

$y=e^{4x}$ is a solution of $y^{\text{'}}=-\frac{1}{4}{y}^2$ that is not a member of the famly in part $($b$).$

$y=4$ is a solution of $y^{\text{'}}=-\frac{1}{4}{y}^2$ that is not a member of the family in part $($b$).$

Every solution of $y^{\text{'}}=-\frac{1}{4}{y}^2$ is a member of the family in part $($b$).$

$y=0$ is a solution of $y^{\text{'}}=-\frac{1}{4}{y}^2$ that is not a member of the family in part $($b$).$