What is the particular solution of the differential equation # y''' + 3y''-4y'-12y =0 #?

The solution is # y = Ae^(-2x) + Be^(2x) + Ce^(-3x)#

1 Answer
Steve M
Jun 22, 2017

We have:

# y''' + 3y''-4y'-12y =0 # ..... [A]

This is a Third order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives.

Complimentary Function

The Auxiliary equation associated with the homogeneous equation of [A] is:

# m^3 +3m^2-4m-12 = 0 #

The challenge with higher order Differential Equation is solving the associated higher order Axillary equation. By inspection we see #m=2# is a solution, so we can factor out #(x-2)# and use polynomial division to get a quadratic factor:

# (m-2)(m^2+5m+6) = 0#
# :. (m-2)(m+2)(m+3) = 0#

Which has three real distinct solution #m=+-2,-3#.

The roots of the axillary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

  • Real distinct roots #m=alpha,beta, ...# will yield linearly independent solutions of the form #y_1=Ae^(alphax)#, #y_2=Be^(betax)#, ...
  • Real repeated roots #m=alpha#, will yield a solution of the form #y=(Ax+B)e^(alphax)# where the polynomial has the same degree as the repeat.
  • Complex roots (which must occur as conjugate pairs) #m=p+-qi# will yield a pairs linearly independent solutions of the form #y=e^(px)(Acos(qx)+Bsin(qx))#

Thus the solution of the homogeneous equation is:

# y = Ae^(-2x) + Be^(2x) + Ce^(-3x)#

Note this solution has #3# constants of integration and #3# linearly independent solutions, hence their superposition is the General Solution.

This is consistent with the suggested solution.

Related questions
Impact of this question
6985 views around the world
You can reuse this answer
Creative Commons License