The tutor shows the quadratic method (pka, characteristic equation) behind 2nd order LDEs w/ constant coefficients.

An exponential function of the form

y=e^kt

has the first derivative

y’=ke^kt

and second derivative

y”=k²e^kt

Seeing an equation like, for example,

y” -3y’ +2y = 0

one can suppose the solution to be of the form

y=Ce^kt (C is some constant)

so the equation becomes

k²Ce^kt -3kCe^kt +2Ce^kt = 0

Now we can factor out Ce^kt:

Ce^kt(k² -3k +2)=0

and then factor and solve for k:

Ce^kt(k-2)(k-1)=0

k=2 or k=1

Therefore, a possible solution to the equation

y” -3y’ +2y = 0

is

y=C₁e^1t + C₂e^2t

or just

y=C₁e^t + C₂e^2t

HTH:)

Source:

Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems. New York: John Wiley & Sons, Inc., 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.