$\displaystyle \frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}$

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?

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- Feb 10th 2009, 07:11 AMbutbi9xMOD can delete this topic
$\displaystyle \frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}$

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? - Feb 11th 2009, 10:28 AMMoo
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, $\displaystyle \frac{dy}{dt}<0$