Constant-Coefficient Homogeneous ODE

A constant-coefficient homogeneous second-order ode can be
put in the form

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where p and q are constants. Recall that the general solution is

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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

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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

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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:

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Consider the following example:

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The characteristic polynomial is r^2 - 3r -18 = (r - 6)(r +3), which
has roots 6 and -3.

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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

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Consider the following example:

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The characteristic polynomial is r^2 + 6r + 9 = (r + 3)^2, which has a
double root -3. The general solution is

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Complex-Conjugate Roots

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:

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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:

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Using this identity, we have:

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Similarly, we have

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Substituting, these two expressions into the general solution we have

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Choose C_1=0.5 and C_2=0.5. This yields the solution:

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Now choose C_1=i0.5 and C_2=-i0.5. This yields the solution:

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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:

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where D_1 and D_2 are constants.

Consider the example:

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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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