Section 3.6: Nonhomogeneous 2nd Order D.E.’s
Method of Undetermined Coefficients
Christopher Bullard
MTH-314-001
5/12/2006
Nonhomogeneous Equations: Assumptions
• Form: L(y)= y’’ + p(t)y’ + q(t)y = g(t), where g(t) is not
equal to zero.
• p(t) and q(t) are continuous for all t in the domain.
Now consider the homogeneous equation
y’’ + p(t)y’ + q(t)y =0
If we consider the only difference between the
equations as their end result, g(t), we can claim…
Theorem 3.6.1
If Y1 and Y2 are solutions to L(y), then Y1 - Y2 is a solution
to the homogeneous equivalent of L(y) (where g(t) = 0).
This means if y1 and y2 are solutions to the
homogeneous equation, then
Y1 - Y2 = c1y1(t) + c2y2(t)
How does this help us solve the nonhomogeneous
equation? Consider the following proof…
Proof: Assume Y1 and Y2 meet the following requirements:
L[Y1](t) = g(t) and L[Y2](t) = g(t)
Subtracting the two, we find…
L[Y1](t) - L[Y2](t) = g(t) – g(t) = 0
Using the distributive property…
L[Y1(t) - Y2(t)]= L[Y1](t) - L[Y2](t)
Therefore, L[Y1(t) - Y2(t)]= c1y1(t) + c2y2(t),
which leads us to our next theorem…
Theorem 3.6.2:
A solution to the nonhomogeneous equation L(y) is of the
form…
y(t)= c1y1(t) + c2y2(t) + Y(t)
Remember c1y1(t) + c2y2(t) are the constants and solutions
to the homogeneous equation. Y(t) is the
nonhomogeneous solution, while y(t) is the complete
solution to the general nonhomogeneous equation.
Steps to find the solution Y(t):
• Find c1, y1, c2, and y2.
• Find a particular solution Y(t) to the nonhomogeneous
equation (anything that solves y’’ + p(t)y’ + q(t)y = g(t)).
• Add Y(t) to c1y1(t) + c2y2(t), and then solve for the initial
conditions.
But where do undetermined coefficients come into play?
Undetermined Coefficients: Uses
Advantages:
•Very easy procedure
(assuming a good guess)
Disadvantages:
•Difficult to guess Y(t)
•Limitations on the number of
equations we can solve based
on the form of g(t).
So, which forms of g(t) can we solve for with the Undetermined
Coefficients Method?
•Polynomials
•Exponential (ert) functions
•Trigonometric functions with sin(x) and cos(x)
Potential Problems:
• If Y(t) results in the same solution to the homogeneous
equation (where g(t) = 0), we need to find another
solution. Usually, the answer is to multiply Y(t) by t. If this
does not work, multiply by t again (to get t2 attached to
Y(t)).
• Why does this work? From Section 2.1, we could solve
first order differential equations with a similar problem (y’
– y = 2e-t). We find y(t) in this case was 2te-t. We just use
this similar case as a template for our approach to cases
when Y(t) = y1 or y2.
Summary of Procedures:
• Find the homogeneous solution c1y1(t) + c2y2(t).
• If g(t) is a polynomial, exponential, or trigonometric
function with sin(x) or cos(x), we can find Y(t).
• What about combinations of the above? Simply find Y(t)
for the individual cases, and algebraically add them
together.
• Add the homogeneous and nonhomogeneous solutions.
• Use the initial conditions to find the particular solution.
Other Examples and Sources of Information:
• S.O.S. Math: Method of Undetermined Coefficients
• Paul's Online Math Notes (Lamar University)
• Differential Equations Notes ( Bruce Ikenaga of
Millersville University)
Work Cited:
Boyce, William E., DiPrima, Richard C. Elementary Differential
Equations: Eighth Edition. New York:John Wiley and Sons, Inc. 2005