A constant-coefficient homogeneous second-order ode can be
put
in the form
where p and q are constants. Recall that the general solution is
where C_1 and C_2 are constants and y_1(t) and y_2(t) are any
two
linearly independent solutions of the ode. Our goal is
to find two linearly
independent solutions of the ode.
For reasons that become clear below, we try a solution of the
form
y=exp(rt), where r is an unknown constant. We must find r.
If y=exp(rt),
then y'=r(exp(rt)) and y''=r^2(exp(rt)).
Substituting these into
the ode, we have
This equation is satisfied if exp(rt)=0 or r^2+pr+q=0.
The condition
exp(rt)=0 is satisfied only if rt is negative
infinity. This is considered a
degenerate case and is
neglected. Hence, r must satisfy the equation
F(r) is called the characteristic polynomial. Solving the
original ode is
reduced to solving an algebraic equation.
Assuming the coefficients p and q are real numbers, there
are three
cases to consider:
Characteristic Polynomial has Distinct Roots
Suppose that the characteristic polynomial has two distinct,
real roots
(call them a and b). Then exp(at) and exp(bt) are
linearly independent
solutions to the original ode and
the general solution to the ode is:
Consider the following example:
The characteristic polynomial is r^2 - 3r -18 = (r - 6)(r +3),
which
has roots 6 and -3.
Characteristic Polynomial has a Double Root
Suppose that the characteristic polynomial has a double root
(call it a).
y_1=exp(at) is one solution to differential equation.
We need a second
linearly independent solution to the ode to
get the general solution. Using
the technique of
reduction of
order, it can be shown that texp(at) is also
a solution of the
ode. The general solution is
Consider the following example:
The characteristic polynomial is r^2 + 6r + 9 = (r + 3)^2,
which has a
double root -3. The general solution is
Suppose that the characteristic polynomial has complex
roots a+ib and
a-ib, where a and b are real. These are distinct
roots, so the the general
solution can be written:
The problem with writing the solution in this form is
that it involves
complex-valued functions. It is possible
to re-express the general
solution in terms of two linearly
independent real-valued functions.
Recall, Euler's identity:
Using this identity, we have:
Similarly, we have
Substituting, these two expressions into the general solution we have
Choose C_1=0.5 and C_2=0.5. This yields the solution:
Now choose C_1=i0.5 and C_2=-i0.5. This yields the solution:
Both of these functions are solutions to the original ode. In
addition, they
are linearly independent, since they are not
multiples of each. Hence, any
solution to the ode can be
expressed in terms of these function. So we
can write:
where D_1 and D_2 are constants.
Consider the example:
The characteristic polynomial is r^2-6r+13. Using the
quadratic formula,
we find that the roots are 3 + 2i and 3 - 2i.
Hence, the general solution
can be written:
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