Answer to Question #231182 in Differential Equations for Randal Rodriguez

Question

determine the values of r for which the given differential equation has solutions of the form y = e^rt

11.y′′′\:−\:3y′′\:+\:2y′\:=\:0

Answer 1

$\text{The auxilliary equation for the given differential equation is}\\ y^3-3y^2+2y=0\\ \text{Next, we factorise the auxillary equation, since $y-1$ is a factor, we have that}\\ (y^2 + ay + b)(y-1) = y^3 -3y^2 +2y\\ =y^3 +y^2(a-1) +y(b-a) = y^3 -3y^2 +2y\\ \text{Comparing co-efficients, we have that}\\ a-1 = -3 \therefore a=-2\\ b+2 = 2 \therefore b = 0\\ \implies (y^2-2y)(y-1)=0\\ =y(y-1)(y-2)=0\\ \text{Hence $y =0$, $y = 1$ and $y=2$, we know that the solution of the differential equation is}\\ x = ce^{yt}\\ \therefore \text{the values of r are 0, 1 and 2}$