$\displaystyle \frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}$
a. What are the constant solutions of the equation?
b. For what values of x is y increasing?
c. For what values of x is y decreasing?
$\displaystyle \frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}$
a. What are the constant solutions of the equation?
b. For what values of x is y increasing?
c. For what values of x is y decreasing?
Should be t (in red)
a. If y=a, where a is a constant, then $\displaystyle \frac{dy}{dt}=0$
So solve for a in :
$\displaystyle a^4-6a^3+5a^2=0$
$\displaystyle 0=a^2(a^2-6a+5)=a^2(a-1)(a-5)$
So y=0, y=1 and y=5 are the constant solutions to the equations.
b. If y is increasing, it means that $\displaystyle \frac{dy}{dt}>0$
So find the values (I guess they should still be constant), such that $\displaystyle y^4-6y^3+5y^2>0 \Leftrightarrow y^2-6y+5>0$
Same thing for c. : if y is decreasing, $\displaystyle \frac{dy}{dt}<0$