LECTURE 14: NONHOMOGENEOUS EQUATIONS
MINGFENG ZHAO
October 14, 2015
Theorem 1. Let yc (x) be the general solution to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 ( which is call the
complementary solution), and yp (x) be any particular solution to the nonhomogeneous equation y 00 +p(x)y 0 +q(x)y = f (x),
then the general solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x) is:
y(x) = yp (x) + yc (x).
Proof. Let y be any solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x), then
y 00 + p(x)y 0 + q(x)y = f (x).
Let yp (x) be any particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x), then
yp00 + p(x)yp0 + q(x)yp = f (x).
Let yc (x) = y(x) − yp (x), then we know that yc00 + p(x)yc0 + q(x)yc = 0, that is, yc is a solution to the homogeneous
equation y 00 + p(x)y 0 + q(x)y = 0.
Undetermined Coefficients:
Let a, b and c be constants, consider the equation:
ay 00 + by 0 + cy = f (x).
Let pn (x) and p̃n (x) be polynomials with degree n, a particular solution yp (x) to ay 00 + by 0 + cy = f (x) can be taken as:
f (x)
yp (x)
pn (x)emx cos(kx) + p̃n (x)emx sin(kx)
xα [qn (x)emx cos(kx) + q̃n (x)emx sin(kx)]
where
1
2
MINGFENG ZHAO
• α is one of 0, 1 and 2 (α is the multiplicity of m+ki as the solutions to the characteristic equation ar2 +br +c = 0
of ay 00 + by 0 + cy = 0):
– If m + ki is not a root of ar2 + br + c = 0, then α = 0.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 has two different roots, then α = 1.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 only has one root m + ki, then α = 2.
• qn (x) and q̃n (x) are undetermined polynomials with degree n.
Example 1. Find the general solution to y 00 + 5y 0 + 6y = 2x + 1.
The characteristic equation to y 00 + 5y 0 + 6y = 0 is :
r2 + 5r + 6 = 0.
Solve r2 + 5r + 6 = 0, we get
r1 = −2,
and r2 = −3.
So the general solution to the homogeneous equation y 00 + 5y 0 + 6y = 0 is:
y(x) = C1 e−2x + C2 e−3x .
Let yp (x) = Ax + B be a particular solution to y 00 + 5y 0 + 6y = 2x + 1, then
yp00 + 5yp0 + 6yp = 0 + 5A + 6(Ax + B) = 2x + 1.
That is, we have
(6A − 2)x + 5A + 6B − 1 = 0.
So
6A − 2 = 0,
and
5A + 6B − 1 = 0.
So
A=
1
,
3
1
and B = − .
9
x 1
− is a particular solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1. Therefore, he
3 9
general solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1 is:
Hence yp (x) =
y(x) =
x 1
− + C1 e−2x + C2 e−3x .
3 9
LECTURE 14: NONHOMOGENEOUS EQUATIONS
3
Example 2. Find a particular solution to y 00 + 2y 0 + 3y = e5x .
The characteristic equation to y 00 + 2y 0 + 3y = 0 is :
r2 + 2r + 3 = 0.
Solve r2 + 5r + 6 = 0, we get
√
r1 = −1 + i 2,
√
and r2 = −1 − i 2.
Let yp (x) = Ae5x be a particular solution to y 00 + 2y 0 + 3y = e5x , then
So 38A = 1, that is, A =
1
38 .
yp0 (x)
=
5Ae5x
yp00 (x)
=
25Ae5x
yp00 + 2yp0 + 3yp
=
25Ae5x + 10Ae5x + 3Ae5x
=
38Ae5x
=
e5x
Hence a particular solution to y 00 + 2y 0 + 3y = e5x is yp (x) =
1 5x
e .
38
Example 3. Find a particular solution to y 00 − 6y 0 + 9y = e3x .
The characteristic equation of y 00 − 6y 0 + 9y = 0 is:
r2 − 6r + 9 = 0.
Solve r2 − 6r + 9 = 0, we get
r1 = r2 = 3.
Let yp (x) = Ax2 e3x be a particular solution to y 00 − 6y 0 + 9y = e3x , then
yp0 (x)
=
2Axe3x + 3Ax2 e3x
yp00 (x)
=
2Ae3x + 6Axe3x + 6Axe3x + 9Ax2 e3x
=
2Ae3x + 12Axe3x + 9Ax2 e3x
=
2Ae3x + 12Axe3x + 9Ax2 e3x − 6 2Axe3x + 3Ax2 e3x + 9Ax2 e3x
=
2Ae3x
=
e3x .
yp00 − 6yp0 + 9yp
4
MINGFENG ZHAO
So we have 2A = 1. That is, A =
1
1
. So a particular solution to y 00 − 9y = e3x is yp (x) = x2 e3x .
2
2
Example 4. Find a particular solution to y 00 + y = cos(2x).
The characteristic equation to y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
r1 = i,
and r2 = −i
Let yp (x) = A cos(2x) + B sin(2x) be a particular solution to y 00 + 2y 0 + 2y = cos(2x), then
yp0 (x)
=
−2A sin(2x) + 2B cos(2x)
yp00 (x)
=
−4A cos(2x) − 4B sin(2x)
yp00 + yp
=
−4A cos(2x) − 4B sin(2x) + A cos(2x) + B sin(2x)
=
−3A cos(2x) − 3B sin(2x)
=
cos(2x).
So
−3A = 1,
and
− 3B = 0.
So we have
1
A=− ,
3
and B = 0.
1
So a particular solution to y 00 + y = cos(2x) is yp (x) = − cos(2x) .
3
Example 5. Find a particular solution to y 00 + y = cos(x).
The characteristic equation to y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
r1 = i,
and r2 = −i
Let yp (x) = x[A cos(x) + B sin(x)] be a particular solution to y 00 + y = cos(x), then
yp0 (x)
= A cos(x) + B sin(x) + x[−A sin(x) + B cos(x)]
yp00 (x)
= −A sin(x) + B cos(x) − A sin(x) + B cos(x) + x[−A cos(x) − B sin(x)]
LECTURE 14: NONHOMOGENEOUS EQUATIONS
5
= −2A sin(x) + 2B cos(x) − x[A cos(x) + B sin(x)]
yp00 + yp
= −2A sin(x) + 2B cos(x) − x[A cos(x) + B sin(x)] + x[A cos(x) + B sin(x)]
= −2A sin(x) + 2B cos(x)
=
cos(x).
So
−2A = 0,
and
2B = 1.
So we have
A = 0,
So a particular solution to y 00 + y = cos(x) is yp (x) =
and B =
1
.
2
x sin(x)
.
2
Problems you can do:
Lebl’s Book [2]: Exercise 2.5.2, 2.5.3, 2.5.4, 2.5.7 b), 2.5.9, 2.5.101, 2.5.102, 2.5.103 and 2.5.105 on Page 75
and Page 76.
Braun’s Book [1]: All exercises on Page 164. Read all materials in Section 2.5.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca