# 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”=k^{2}e^{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

k^{2}Ce^{kt} -3kCe^{kt} +2Ce^{kt} = 0

Now we can factor out Ce^{kt}:

Ce^{kt}(k^{2} -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=C_{1}e^{1t} + C_{2}e^{2t}

or just

y=C_{1}e^{t} + C_{2}e^{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.