$($a$)$ What can you say about a solution of the equation $y^{\text{'}}=-\frac{1}{6}{y}^2$ Just by looking at the afferencial equation?

The function y must be increasing for eoual so $01$ on affy interval on which it is deflede.

The function y most be strictly deoreasing en any interval on which is is defined.

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

The function y must be decreasing $($or equal to $0)$ on aty interval en which it is def ned.

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

$($b$)$ Verify that all members of the famlly $y=\frac{6}{(xC)}$ are solutions of the equation in part $($a$).$

We substitute the values of $y$ and $y^{\text{'}}$ and best the solution tes see if the left hand slope $($L$.$HS$)$ it equal to the right hand slide $($RAS$).$

$y=\frac{6}{xC}-y^2=-\frac{?}{(xC{)}^2}$

thes$=y^{\text{'}}=-\frac{?}{(xC{)}^2}=-\frac{1}{6}(\frac{}{xC}{)}^2=-\frac{1}{6}{r}^2=sens$

$($c$)$ Can you think of a solution of the differential equation $y^{\text{'}}=-\frac{1}{6}{y}^2$ that is not a member of the family in part $($b$)?$

$=0$ is a solution of $y^2=-\frac{1}{6}{y}^2$ that is not s momere at the fomity in part $($b$).$

Every solvition of $y^{\text{'}}=-\frac{1}{6}{y}^2$ is a member of the family in part $($b$).$

$y=x$ is a solvition of $y^{\text{'}\text{'}}=-\frac{1}{6}{y}^2$ that is not a member of the family in part $($b$).$