6. Constant Coefficient, Homogeneous ODEs

The focus in this section of the course is the solution of constant coefficient, homogeneous, LINEAR ODEs.  The limitation that the ODE must be linear arises since we will find fundamental solutions to the ODE (e.g. y1 and y2) and then use the principle of superpositiong to form the general solution (provided that y1 and y2 are linearly independent).  For example, the general solution to a second order linear ODE  is given by superposition of the linearly independent solutions y1 and y2.  The linear combination is:
    yGeneral = C1*y1 + C2*y2

where, like always, C1 and C2 are determined using the boundary/initial conditions.

Solution strategies/ideas will be presented in the context of 2nd order ODEs, and then generalized to higher order ODEs.  Note, however, that the constaints of constant coefficient, homogenous and linear still apply!

Some notational observations for second order ODEs before we get started:

  • Get the ODE into standard form.  E.g.  d2y/dx^2 + a1*dy/dx + a2*y= 0
  • Other notations include:
  • y'' + a1*y' + a2*y = 0
  • Dot notation for time derivatives (hard to type them with normal text)
  • (D^2 + a1*D + a2)y = 0, where D represents the differential operator d/dx
  • phi(D)y = 0, where phi(D) is shorthand for the  (D^2 + a1*D + a2)  operator only
    • While the first three notations generalized to higher than second order ODEs, phi(D) does not; it refers to the second order ODE only.  This is presumably due to the commonality of this ODE in engineering applications.

    The complete solution process has only a few steps:
    1.    Assume a fundamental solution form: y=exp(m*x) always works for C.C., Homog, linear ODEs, where m is a constant.
    2.    Substitute this into the ODE
    3.    Determine the Characteristic/Auxiliary Equation
    4.    Use its roots to find additional fundamental solutions, and form the fundamental set
            NOTE: You will need to use D'Alembert in the case of repeated roots (see below)
    5.    Form the general solution as a linear combination of the fundamental solutions
    6.     Apply IC/BC information to obtain the specific solution

    6.1.   The Characteristic/Auxiliary Equation

    Here, we consider the first three steps in the solution process.
    First assume that a fundamental solution exists with the form: y = exp(m*x).... we need to find m
    Next, find the derivatives required to substitute this soln. into the ODE:
        y'     = m*exp(m*x)
        y''    = m^2 * exp(m*x)

    Substitute them into the ODE, and find that the exp(m*x) factors cancel, leaving only a polynomial equation - the auxiliary or characteric equation -  in the constants a1 and a2 (more for higher order ODEs) and the unknown, m.

    The roots of this polynomial equation (i.e. values of m for which the polynomial is satisfied) are the values of m that will give us some, if not all, of the fundamental solutions to the ODE.

    Before looking at the three cases of roots which need to be considerred (real distinct roots, repeated roots, complex roots), it is important to reiterate that a second order ODE will lead to a second order characteristic equation with 2 roots.  A third order ODE will lead to a third order characterist equation with 3 roots....

    6.2.    Real and distinct roots

    Perhaps the simplest case to consider involves real and distinct roots.  A full sample solution is given here for the problem:
        y'' - 3*y' + 2*y = 0,     with y(0)=1, and y'(0)=0

    6.3.    Repeated Roots

    Next, we may observe that the roots of the characteristic equation may be the same. In these cases, only one fundamental solution arises from the root(s) - i.e. y1(x), and the other must be found using D'Alembert's method.

    Application of D'Alemberts method reveals that the y2=v(x)*y1, where
        v(x) = A*x + B

    That is to say, y2(x) = (A*x+B)*y1(x).  This result says that y2 will be LI of y1, for ANY values of A and B that preserve the the functional dependence (linear in this case) of v(x) on x... in the end, we choose the simplest straight line for v(x).... let A=1 and B=0 (NOTE: A=0 is NOT acceptable, since it gets rid of the x).  The resulting fundamental set of solutions is:
        y1(x)
        y2(x) = x*y1(x)

    This is for an ODE with 2 repeated roots.  Indeed, the solution form may be generalized to ODEs with j repeated roots.  The LI solution corresponding to the jth repeated root is simply:
        y_j(x) = x^j * y1(x)

    where y1(x) is the original root found using the characteristic equation.

    Click here for an example with two roots, and see below for an example solution of a 14th order ODE!!!

    6.4.    Complex Roots

    Finally, the potential for complex roots also needs to be accounted for.  The Euler identity (relating complex exponentials to sines and cosines) is used in this case.  Indeed, the final solution y(x) is independent of complex numbers.

    The general result that follows from the investigation of ODEs that have chacteristic equations with complex roots:
        m = a +- i*b       (where i is sqrt(-1))

    there are a couple of solution forms for the two fundamental solutions:
    1)    y1 = exp((a+i*b)*x)               y2 = exp((a-i*b)*x)
    2)    y1=exp(a*x) * cos(b*x)        y2 = exp(a*x) * sin(b*x)

    and the general solution is:
            y = exp(a*x) * ( c1*cos(b*x) + c2*sin(b*x) )
    or,
            y = exp(a*x) * K * sin(b*x + phi)
            where     K = sqrt(c1^2+c2^2)
                            phi = atan(c1/c2)

    A couple of examples are given here, in addition some more of the details regarding how to get between solution forms.

    6.5.    Putting It All Together

    How about this 14th order ODE....   (click here)