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Find and classify all the equilibrium solutions to the subsequent differential equation.
y' = y2 - y - 6
Solution
First, get the equilibrium solutions. It is generally easy adequate to do.
y2 - y - 6 = (y - 3) ( y + 2) = 0
Thus, it looks as we've found two equilibrium solutions. y = -2 and y = 3 both are equilibrium solutions. There is the sketch of several integral curves for this differential equation is given below. A sketch of the integral curves or direction fields can simplify the method of classifying the equilibrium solutions.
By this sketch it appears that solutions which start "near" y = -2 all move indirections of it as t rises and so y = -2 is an asymptotically stable equilibrium solution and solutions which start "near" y = 3 all move away from it as t raises and so y = 3 is an unstable equilibrium solution.
This next illustration will introduce the third classification which we can give to equilibrium solutions.
In 6th grade I am learning about ratios rates and fractions. I am working on vmathlive.com and need serious.
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