Answer
A) Decreasing function or $0$
B) $y = \frac{1}{(x+C)}$ is a solution of the equation of $ y' = -y^{2}$
C) $y = 0$
D) $y = \frac{1}{(x+2)}$
Work Step by Step
A) $y' = -y^{2}$
Since $y'$ is negative, the solution of this equation must be a decreasing function or $0$.
B) $y = \frac{1}{(x+C)}$
$=> y' = -\frac{1}{(x+C)^{2}}$
$=> y' = -y^{2}$
Hence, $y = \frac{1}{(x+C)}$ is a solution of the equation of $ y' = -y^{2}$
C) $y = 0$ is a solution of the equation but it is not a member of the family of curves $y = \frac{1}{(x+C)}$
D) The equation is $ y' = -y^{2}$
The solution is $y = \frac{1}{(x+C)}$ trom part (b)
Substitute x = 0 and y = 0.5, since y (0) = 0.5
$\frac{1}{(C)} = 0.5$
$=> C =\frac{1}{0.5}$
$C=2$
Thus, the solution of the equation is $y = \frac{1}{(x+2)}$
