Question

A function y(t) satisfies the differential equation dy dt = y4 − 6y3 + 5y2.

Answer 1

$\dfrac{\mathrm dy}{\mathrm dt}=y^4-6y^3+5y^2=y^2(y-1)(y-5)$

a. If $y(t)$ is constant, then the derivative 0, which means we would have

$y=0\text{ or }y=1\text{ or }y=5$

as constant solutions.

Next, we have 4 possible intervals to consider where the derivative doesn't vanish:

• for $t$, we have $\frac{\mathrm dy}{\mathrm dt}>0$ (consider the sign of the derivative for, say, $y=-1$);
• for $0$, we have $\frac{\mathrm dy}{\mathrm dt}>0$;
• for $1$, we have $\frac{\mathrm dy}{\mathrm dt}$;
• and for $t>5$, we have $\frac{\mathrm dy}{\mathrm dt}>0$

Taking all these facts together, we see that ...

b. $y$ is increasing on the interval $(-\infty,0)\cup(0,1)\cup(5,\infty)$, and

c. $y$ is decreasing on the interval $(1,5)$.