MATH 2280-002
Lecture Notes: 2/06/2013
Math 2280 Section 002 [SPRING 2013]
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Nonhomogeneous Linear DE’s
Until now, we’ve focused on homogeneous linear DE’s, but for us to handle forced (damped or
undamped) motion of mass-spring systems, we need to be able to handle nonhomogeneous linear
DE’s.
Theorem. Suppose yp is a paricular solution to the nonhomogeneous DE
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = f (x)
on an open interval I where p1 , ..., pn , and f are continuous. Let y1 , y2 , ..., yn be linearly independent
solutions of
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = 0.
Then the general solution to the above nonhomogeneous DE is
y(x) = c1 y1 + c2 y2 + ... + cn yn + yp
for all x in I.
First of all, this theorem tells us that the set of solutions to a nonhomogeneous linear DE is an
n-dimensional subspace of the function space, just like the homogeneous case. You should think of
each of the c1 , ..., cn as independent choices, and you should think of the number of independent
choices as being synonymous with dimension. So even though there are n + 1 summands in the
general solution, the solution space is only n-dimensional.
The theorem tells us that if yp is any particular solution to a nonhomogeneous linear DE
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = f (x)
and yh is the general solution to the associated homogeneous linear DE
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = 0,
the general solution to our orginal DE is
y(x) = yh + yp .
The yh part is necessary to satisfy our initial conditions, and the yp part allows us to obtain f (x) when
we plug y(x) into the nonhomogeneous DE. The yp part doesn’t depend on the initial conditions. As
noted above, any particular solution for the DE will work, so we’re lead to ask the following question:
Question. How do we find a particular solution to a nonhomogeneous linear DE?
To find yp for
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = f (x),
we’ll learn two techniques:
1. Undetermined Coefficients. This method works only for some types of f (x). Success is
guaranteed by linear algebra.
2. Variation of Parameters. This method always works but formulae are (very!) messy.
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MATH 2280-002
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Lecture Notes: 2/06/2013
Undetermined Coefficients
IDEA:
If xm , sin(kx), cos(kx), or erx appear in f (x), then they appear in yp .
Roughly speaking, we notice that the only way to get polynomials, erx , sin(kx), cos(kx), or certain products/sums of these functions by integrating or taking the derivative is to start with these
functions.
Undetermined Coefficients Method. If f is of the form
(a0 + a1 x + ... + am xm )erx cos(kx) + (b0 + b1 x + ... + bm xm )erx sin(kx),
take as a trial solution
yp (x) =xs [(A0 + A1 x + ... + Am xm )erx cos(kx)
+ (B0 + B1 x + ... + Bm xm )erx sin(kx)],
where s is the smallest nonegative integer such that no term in yp duplicates a term in yh . Then
determine the coefficients by substituting yp into the original nonhomogeneous equation.
For some sample values of m, r, k, we’ve tabulated below f (x) and the appropriate trial function.
f (x)
Pm (x) = a0 + a1 x + a2 x2 + ... + am xm
(r = 0, k = 0)
a cos(kx) + b sin(kx)
(m = 0, k = 0)
erx (a cos(kx) + b sin(kx))
(m = 0)
Pm (x)erx
(k = 0)
Pm (x)(a cos(kx) + b sin(kx))
(r = 0)
xs (A
yp
2
m
0 + A1 x + A2 x + ... + Am x )
xs (A cos(kx) + B sin(kx))
xs erx (A cos(kx) + B sin(kx))
xs erx (A0 + A1 x + A2 x2 + ... + Am xm )
xs [A cos(kx)(A0 + A1 x + A2 x2 + ... + Am xm )
+B sin(kx)(A0 + A1 x + A2 x2 + ... + Am xm )]
We can use superposition to solve even more DE using this method. The best way to illustrate this
method is to work through some examples.
Example. Find the general solution for
y 0 + 5y = e−5x cos(2x).
Begin by finding the solution to the associated homogeneous DE
y 0 + 5y = 0.
The characteristic polynomial p(r) of this equation has a single, distinct root at r = −5, so
yh (x) = c1 e−5x .
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MATH 2280-002
Lecture Notes: 2/06/2013
The method of undetermined coefficients tells us to take
yp (x) = xs e−5x (A cos(2x) + B sin(2x))
as our trial solution. Since there is no duplication in this case s = 0. By the product rule,
yp0 = −5e−5x (A cos(2x) + B sin(2x)) + e−5x (−2A sin(2x) + 2B cos(2x))
Plug yp into our nonhomogeneous DE to solve for A and B:
yp0 + 5yp = e−5x cos(2x)
−2Ae−5x sin(2x) + 2Be−5x cos(2x) = e−5x cos(2x)
We match equations on the left-hand side with those appearing on the right-hand side and equate
coefficients belonging to similar terms. (We can do this because of the linear independence of these
functions.) When we do this, we get 2B = 1 (or B = 21 ) and A = 0. We can now write down the
desired general solution:
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y(x) = yp (x) + yh (x) = e−5x sin(2x) + c1 e−5x .
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Example. Find the general solution for
y 4 − 2y (3) = 180x2 + 24x.
The characteristic polynomial p(r) of y 4 − 3y (3) = 0 has a distinct root at r = 2 and a repeated
root at r = 0 with multiplicity 3, so
yh (x) = c1 e2x + c2 + c3 x + c4 x2 .
The method of undetermined coefficients tells us to take
yp (x) = xs (Ax2 + Bx + C),
where s = 3 in this case. Plug yp into our original DE to get
yp4 − 3yp(3) = 120Ax + 24B − 180Ax2 − 72Bx − 18C.
Setting this equal to x2 + 5x implies that
−180A = 180, 120A − 72B = 24, 24B − 18C = 0,
so A = −1, B = −2, and C = − 38 . The general solution is
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y(x) = c1 e2x + c2 + c3 x + c4 x2 + −x5 − 2x4 − x3 .
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Example. Find the general solution for
y 00 − 9y = e3x + x.
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MATH 2280-002
Lecture Notes: 2/06/2013
The characteristic polynomial p(r) of y 00 − 9y = 0 has a distinct root at r = 3 and at r = −3, so
yh (x) = c1 e3x + c2 e−3x .
We can’t just apply the method of undetermined coefficients because e3x + x doesn’t match our
requirements. Because of superposition, we notice that adding a particular solution for y 00 −9y = e3x
to a particular solution for y 00 − 9y = x gives us a particular solution for y 00 − 9y = e3x + x. The
method of undetermined coefficients gives us the trial solution Axe3x for y 00 − 9y = e3x and the
trial solution Bx + C for y 00 − 9y = x. That means
yp (x) = Axe3x + Bx + C.
I’ll leave the rest for you to work out because the remainder of the problem can be solved in a
similar manner to the previous two examples. (Plug yp into the original DE to solve for A, B, C,
and then the general solution y(x) = yp (x) + yh (x).)
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