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Homogeneous linear differential equations with constant coefficients, Auxiliary equation, solutions

We shall here treat the problem of finding the general solution to the homogeneous linear differential equation with constant coefficients

Such an equation can be written in the operator form

or, more simply,

f(D)y = 0

where f(D) is a differential operator.

Def. Auxiliary equation. The auxiliary equation of equation 1) is the equation

f(m) = a_{0 }m^{n} + a_{1 }m^{n -1} + ...... + a_{n-1 }m + a_{n} = 0

The function f(m) may be obtained by simply substituting m for D in the differential operator

of equation 2).

Theorem. If m is any root of f(m) = 0 , then

f(D)e^{mx} = 0,

which means that y = e^{mx} is a solution of equation 1).

The auxiliary equation has n roots. Let the n roots be m_{1}, m_{2}, ..... , m_{n}. If these roots are all real
and distinct, then it can be shown that the n solutions

are linearly independent. The general solution is then given by

where c_{1}, c_{2}, ..... , c_{n} are arbitrary constants.

However some roots may not be real and distinct. Some roots may be multiple roots, complex roots or both. The set of all real, distinct roots make one contribution to the general solution. Each set of multiple roots makes its own separate contribution. The complex roots make their own contribution. The general solution consists of the sum of the contributions from these different sets.

Contribution from the set of all real, distinct roots. Suppose there are k real,
distinct roots. Let the k roots be m_{1}, m_{2}, ..... , m_{k}. Then the contribution to the general solution
from these k real, distinct roots is

Example. Suppose there are three real roots 2, 5 and -3. Then the contribution to the general solution from these three real roots would be

y_{1} = c_{1}e^{2x} + c_{2}e^{5x} + c_{3}e^{-3x}

Contribution from multiple (or repeated) roots. Suppose a root with a value m = b occurs with a multiplicity q (i.e. the auxiliary equation gives q equal roots with value m = b). Then the contribution to the general solution from this set of q equal roots is

Example. Suppose there are four equal roots 2, 2, 2, 2. Then the contribution to the general solution from these four equal roots would be

y_{2} = c_{1}e^{2x} + c_{2}xe^{2x} + c_{3}x^{2}e^{2x} + c_{4}x^{3}e^{2x}

Contribution from complex roots. The contribution to the general solution from two conjugate complex roots a + bi and a - bi is

Repeated complex roots lead to solutions analogous to those for repeated real roots. For example, if the roots m = a bi occurred three times, the contribution would be

(c_{1} + c_{2}x + c_{3}x^{2})e^{ax} cos bx + (c_{4} + c_{5}x + c_{6}x^{2})e^{ax} sin bx

Example. Suppose the auxiliary equation had the complex roots 2 + 3i and 2 - 3i. Then the contribution to the general solution from these two roots would be

y_{3} = c_{1}e^{2x} cos 3x + c_{1}e^{2x} sin 3x

General solution. The general solution is given by

y = y_{d} + y_{r} + y_{c}

where y_{d} is the contribution from all the distinct real roots, y_{r} is the contribution from all the
repeated roots, and y_{c} is the contribution from all complex roots.

Summary of Procedure

1. Put differential equation in operator form

2. Form auxiliary equation by replacing the D’s in the operator form of the equation with m.

3. Find the roots of the auxiliary equation (using a numerical procedure if necessary).

4. Find the contributions to the general solution from the distinct, real roots, any repeated roots and any complex roots.

5. The general solution is given by

y = y_{d} + y_{r} + y_{c}

where y_{d} is the contribution from all the distinct real roots, y_{r} is the contribution from all the
repeated roots, and y_{c} is the contribution from all the complex roots.

Example. Solve the equation

(D^{4} - 7D^{3} + 18D^{2} - 20D + 8)y = 0

The auxiliary equation is

m^{4} - 7m^{3} + 18m^{2} - 20m + 8 = 0

which has the roots m = 1, 2, 2, 2. The general solution is

y = c_{1}e^{x} + c_{2}e^{2x } + c_{3 }xe^{2x} + c_{3 }x^{2}e^{2x}

Example. Solve the equation

(D^{3} - 3D^{2} + 9D + 13)y = 0

The auxiliary equation is

m^{3} - 3m^{2} + 9m + 13 = 0

which has the roots m = -1, 2 + 3i, 2 - 3i. The general solution is

y = c_{1}e^{-x} + c_{2}e^{2x} cos 3x + c_{3}e^{2x} sin 3x

References

1. Earl D. Rainville. Elementary Differential Equations

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