A function y(t) satisfies the differential equation dy dt = y4 − 6y3 + 5y2. (a) What are the constant solutions of the equation? (Enter your answers as a comma-separated list.) y = __. (b) For what values of y is y increasing? (Enter your answer in interval notation.) y is in ___.(c) For what values of y is y decreasing? (Enter your answer in interval notation.) y is in ___.

[tex]\dfrac{\mathrm dy}{\mathrm dt}=y^4-6y^3+5y^2=y^2(y-1)(y-5)[/tex]a. If [tex]y(t)[/tex] is constant, then the derivative 0, which means we would have[tex]y=0\text{ or }y=1\text{ or }y=5[/tex]as constant solutions.Next, we have 4 possible intervals to consider where the derivative doesn't vanish:for [tex]t<0[/tex], we have [tex]\frac{\mathrm dy}{\mathrm dt}>0[/tex] (consider the sign of the derivative for, say, [tex]y=-1[/tex]);for [tex]0<t<1[/tex], we have [tex]\frac{\mathrm dy}{\mathrm dt}>0[/tex];for [tex]1<t<5[/tex], we have [tex]\frac{\mathrm dy}{\mathrm dt}<0[/tex];and for [tex]t>5[/tex], we have [tex]\frac{\mathrm dy}{\mathrm dt}>0[/tex]Taking all these facts together, we see that ...b. [tex]y[/tex] is increasing on the interval [tex](-\infty,0)\cup(0,1)\cup(5,\infty)[/tex], andc. [tex]y[/tex] is decreasing on the interval [tex](1,5)[/tex].