Lecture 9: Solving Linear Nonhomogeneous DEs
The Method of Undetermined Coefficients, Part I
Winfried Just
Department of Mathematics, Ohio University
February 10, 2017
Winfried Just, Ohio University
MATH3400, Lecture 9: Undetermined Coefficients I
A motivating example
Consider the DE
y 00 + y = 1
y1 (x) = sin x + 1 is a solution.
y2 (x) = cos x + 1 is another solution.
But the linear combination
y (x) = y1 (x) + y2 (x) = sin x + cos x + 2 is not a solution.
y (x) = 2y1 (x) = 2 sin x + 2 also is not a solution.
Linear combinations of solutions of nonhomogeneous linear DEs
usually are not solutions of the DE.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Superposition Principle for Nonhomogeneous Linear DEs
Consider a nonhomogeneous nth -order DE
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = g (x)
The associated linear homogeneous DE aka complementary DE is
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = 0
Let yc (x) be the the general solution of the complementary DE
on an interval I , aka complementary solution, and let
yp (x) be any particular solution of the nonhomogeneous DE
on the same interval I .
Then the general solution of the nonhomogeneous DE on I
is given by y (x) = yc (x) + yp (x).
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Our motivating example revisited
Consider the DE y 00 + y = 1.
yp (x) = sin x + 1 is a particular solution on (−∞, ∞).
The complementary linear homogeneous DE is
y 00 + y = 0.
Its general solution on (−∞, ∞) is
yc (x) = c1 sin(x) + c2 cos x.
The general solution of y 00 + y = 1 on (−∞, ∞) is
y (x) = yc (x) + yp (x) = c1 sin x + c2 cos x + sin x + 1
= c1 sin x + c2 cos x + 1.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Solving nonhomogeneous Linear DEs step-by-step
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = g (x)
1
Form the associated (complementary) linear homogeneous DE:
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = 0
2
Find the general solution yc (x) of the complementary DE
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = 0
3
Find one particular solution yp (x) of the nonhomogeneous DE
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = g (x)
4
Write the general solution y (x) of the nonhomogeneous DE
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = g (x)
as y(x) = yc (x) + yp (x).
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
The method of undetermined coefficients
Step 3 on the previous slide requires us to find
one particular solution yp (x) of
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = g (x).
This method of undetermined coefficients allows us to do this
when all an , an−1 , . . . , a0 are constant and g (x) is of a nice form.
1
Solve the associated homogeneous DE.
2
Express yp (x) in the same general form as g (x) using
coefficients A, B, C , . . . to be determined later.
3
Differentiate yp (x) in its symbolic form n times.
4
Substitute in the DE and match coefficients.
5
Solve for the yet undetermined coefficients.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 0: g (x) is a constant
The method relies on our ability to identify the form of g (x).
If g (x) = b0 , then write yp (x) = A.
Example 0: For y 00 + 4y = 1 we have g (x) = 1.
The complementary solution is yc (x) = c1 sin 2x + c2 cos 2x.
The general solution is y (x) = c1 sin 2x + c2 cos 2x + A.
Differentiate yp (x) = A twice: yp0 (x) = 0, yp00 (x) = 0.
Substitute in the DE: y 00 + 4y = 0 + 4A = 1.
Solve for the coefficient A: We get A = 14 .
The general solution is y (x) = c1 sin 2x + c2 cos 2x + 14 .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 1: g (x) is a polynomial function
If g (x) = b` x ` + b`−1 x `−1 + · · · + b1 x + b0 , then write
yp (x) = A` x ` + A`−1 x `−1 + · · · + A1 x + A0 .
Example 1: For y 00 − y = 2x 2 − x + 7 we have g (x) = 2x 2 − x + 7.
The complementary solution is yc (x) = c1 e x + c2 e −x .
Write yp (x) = Ax 2 + Bx + C with coefficients A, B, C .
Differentiate yp (x) twice: yp0 (x) = 2Ax + B, yp00 (x) = 2A.
Substitute in the DE: y 00 − y = 2A −Ax 2 −Bx − C = 2x 2 −x + 7.
Match coefficients: −A = 2, −B = −1, 2A − C = 7.
Solve for the coefficients: A = −2, B = 1, C = −11.
The general solution is y (x) = c1 e x + c2 e −x − 2x 2 + x − 11.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 1: g (x) is a polynomial function
If g (x) = b` x ` + b`−1 x `−1 + · · · + b1 x + b0 , then write
yp (x) = A` x ` + A`−1 x `−1 + · · · + A1 x + A0 .
Example 2: For y 000 + 3y 00 + 3y 0 + y = x 3 we have g (x) = x 3 .
The complementary solution is yc (x) = c1 e −x + c2 xe −x + c3 x 2 e −x .
Write yp (x) = Ax 3 + Bx 2 + Cx + D with coefficients A, B, C , D.
Differentiate yp (x):
yp0 (x) = 3Ax 2 + 2Bx + C , yp00 (x) = 6Ax + 2B, y 000 = 6A.
Substitute in the DE: y 000 + 3y 00 + 3y 0 + y =
6A +18Ax + 6B + 9Ax 2 + 6Bx + 3C + Ax3 + Bx 2 + Cx + D = x3 .
Match coefficients:
A = 1, 9A + B = 0, 18A + 6B + C = 0, 6A + 6B + 3C + D = 0.
Homework 16: Complete the example by solving for A, B, C , D
and writing the general solution. Convince yourself that setting
yp = Ax 3 would not work here.
Just, Ohio University
Ohio University – Since Winfried
1804
MATH3400, Lecture 9: Undetermined
Coefficients
I
Department of
Mathematics
Case 2: g (x) is an exponential function
If g (x) = e rx , then write yp (x) = Ae rx .
Example 3: For y 00 − 4y =
1
ex
we have g (x) = e −x .
The complementary solution is yc (x) = c1 e 2x + c2 e −2x .
Write yp (x) = Ae −x with coefficient A.
Differentiate yp (x) twice: yp0 (x) = −Ae −x , yp00 (x) = Ae −x .
Substitute in the DE: y 00 − 4y = Ae −x − 4Ae −x = e −x .
Match coefficients: −3A = 1.
Solve for the coefficient: A = − 31 .
The general solution is y (x) = c1 e 2x + c2 e −2x − 31 e −x .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 2: g (x) is an exponential function
If g (x) = e rx , then write yp (x) = Ae rx .
Example 4: For y 00 − 4y = 3x we have g (x) = 3x .
The complementary solution is yc (x) = c1 e 2x + c2 e −2x .
Write yp (x) = A3x = Ae (ln 3)x with coefficient A.
Homework 17: Complete Example 4 by finding the general
solution.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 3: g (x) is a product of a polynomial and an
exponential function
If g (x) = (b` x ` + b`−1 x `−1 + · · · + b1 x + b0 )e rx , then write
yp (x) = (A` x k + A`−1 x `−1 + · · · + A1 x + A0 )e rx .
Example 5: For y 00 − 4y =
x 2 −x
ex
we have g (x) = x 2 e −x − xe −x .
We write yp (x) = (Ax 2 + Bx + C )e −x = Ax 2 e −x + Bxe −x + Ce −x .
Homework 18: Complete Example 5 by finding the general
solution.
Example 6: For y 00 − 4y = x3x we have g (x) = x3x .
We write yp (x) = (Ax + B)3x = Ax3x + B3x .
Homework 19: Complete Example 6 by finding the general
solution.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 4: g (x) is linear combination of sine and cosine
functions
If g (x) = b sin αx + c cos αx, write yp (x) = A sin αx + B cos αx.
Example 7: Consider the DE y 00 − y 0 = sin 3x − cos 3x.
The complementary solution is yc (x) = c1 + c2 e x .
Write yp (x) = A sin 3x + B cos 3x.
Differentiate yp (x) twice:
yp0 (x) = 3A cos 3x − 3B sin 3x, yp00 (x) = −9A sin 3x − 9B cos 3x.
Substitute in the DE: y 00 − y 0 =
−9A sin 3x − 9B cos 3x − 3A cos 3x + 3B sin 3x = sin 3x − cos 3x.
Match coefficients: −9A + 3B = 1, −9B − 3A = −1.
1
Solve for A and B: A = − 15
, B=
2
15 .
The general solution is y (x) = c1 + c2 e x −
Ohio University – Since 1804
Winfried Just, Ohio University
1
15
sin 3x +
2
15
cos 3x.
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 4: g (x) is linear combination of sine and cosine
functions
If g (x) = b sin αx + c cos αx, write yp (x) = A sin αx + B cos αx.
Example 8: For y 00 − y 0 = sin x we have g (x) = sin x.
Although a term c cos x does not appear in g (x), we still need to
include a term B cos x when we set up the form of the particular
solution:
yp (x) = A sin x + B cos x.
Homework 20: Complete Example 8 by finding the general
solution and convince yourself that setting yp (x) = A sin x would
not work.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 4: g (x) is product of an exponential and a sine or
cosine function
If g (x) = be rx sin βx or g (x) = be rx cos βx,
then write yp (x) = Ae rx sin βx + Be rx cos βx.
Example 9: For y 00 + y = e2x sin 3x + 4e2x cos 3x
write yp (x) = Ae 2x sin 3x + Be 2x cos 3x.
Homework 21: Complete Example 9 by finding the general
solution.
Example 10: Consider the DE y 00 + y 0 =
cos x
ex .
We need to write yp (x) = Ae −x sin x + Be −x cos x.
Homework 22: Complete Example 10 by finding the general
solution.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I
Case 5: g (x) contains functions of several forms
If g (x) is a linear combination of functions of several of the above
forms, then add the respective expressions, using a new symbol for
the undetermined coefficient every time.
Example 11: Consider the DE y 00 − 2y 0 + y = cos 2x + xe5x .
Write yp (x) = A sin 2x + B cos 2x + Cxe 5x + De 5x .
Example 12: For the DE y 00 − y 0 = e 2x sin 3x + 4 cos 3x − sin x
Set up yp (x) in the following form:
Ae 2x sin 3x + Be 2x cos 3x + C sin 3x + D cos 3x + E sin x + F cos x.
Homework 23: For each of the following DEs, find and
appropriate form of yp (x).
(a) y 000 − y 0 = 6x +
x
e 4x
− 5x .
(b) y 000 − y 0 = 3x 2 cos 2x + 2xe 3x − e x sin 4x.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 9: Undetermined Coefficients I