LECTURE 14: NONHOMOGENEOUS EQUATIONS
MINGFENG ZHAO
October 06, 2014
Theorem 1. Let y1 (x) and y2 (x) be a fundamental set of solutions to the homogeneous equation y 00 +p(x)y 0 +q(x)y = 0,
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) = C1 y1 (x) + C2 y2 (x) + yp (x).
To find a particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x), we have two methods:
I. Undetermined coefficients: Only work for constant coefficients equations with f (x) are polynomials, exponential, sines, cosines.
II. Variation of parameters: Work for all second order differential equation with all nonhomogeneous term f (x).
Undetermined Coefficients
Let a, b and c be constants, consider the equation:
ay 00 + by 0 + cy = f (x),
The characteristic equation is:
ar2 + br + c = 0.
Then
r1,2 =
−b ±
√
b2 − 4ac
.
2a
For special f (x), we can find a particular solution to ay 00 + by 0 + cy = f (x) by using the method of Undetermined
Coefficients:
I. When c 6= 0 and f (x) is a polynomial, a particular solution can be also a polynomial with the same degree as
f (x).
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MINGFENG ZHAO
II. When f (x) = menx is an exponential function with n 6= r1 and n 6= r2 , a particular solution can be also an
exponential function like yp (x) = Aenx for some constant A.
III. When b2 − 4ac > 0 (that is, r1 and r2 are two different real roots), and f (x) = mer1 x is an exponential function,
a particular solution can be also an exponential function like yp (x) = Axer1 x for some constant A.
Example 1. Find a particular solution to y 00 − 9y = e3x .
The characteristic equation of y 00 − 9y = 0 is:
r2 − 9 = 0.
Solve r2 − 9 = 0, we get
r1 = 3,
and r2 = −3.
Let yp (x) = Axe3x be a particular solution to y 00 − 9y = e3x , then
yp0 (x)
= Ae3x + 3Axe3x
yp00 (x)
=
3Ae3x + 3Ae3x + 9Axe3x
=
6Ae3x + 9Axe3x
=
6Ae3x + 9Axe3x − 9Axe3x
=
6Ae3x
yp00 (x) − 9yp (x)
= e3x
So we have
6A = 1
That is, A =
1
. So a particular solution to y 00 − 9y = e3x is:
6
yp (x) =
1 3x
xe .
6
b
b
), and f (x) = me− 2a ·x is an exponential function, a particular
2a
solution can be also an exponential function like yp (x) = Ax2 erx for some constant A.
IV. When b2 − 4ac = 0 (that is, r1 = r2 = −
Example 2. Find a particular solution to y 00 − 6y 0 + 9y = e3x .
LECTURE 14: NONHOMOGENEOUS EQUATIONS
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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
So we have
2A = 1.
That is, A =
1
. So a particular solution to y 00 − 9y = e3x is:
2
yp (x) =
1 2 3x
x e .
2
V. When f (x) = m sin(kx) + n cos(kx) is a combination of sines and cosines, and ki is NOT a solution to ar2 +
br + c = 0, a particular solution can be also a combination of sines and cosines like yp (x) = A sin(kx) + B cos(kx)
for some constants A and B.
Example 3. Find a particular solution to y 00 + 2y 0 + 2y = cos(2x).
Let yp (x) = A sin(2x) + B cos(2x) be a particular solution to y 00 + 2y 0 + 2y = cos(2x), then
yp0 (x)
=
2A cos(2x) − 2B sin(2x)
yp00 (x)
=
−4A sin(2x) − 4B cos(2x)
yp00 + 2yp0 + 3yp
=
−4A sin(2x) − 4B cos(2x) + 4A cos(2x) − 4B sin(2x) + 2A sin(2x) + 2B cos(2x)
=
cos(2x).
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MINGFENG ZHAO
That is,
[−4B + 4A + 2B − 1] cos(2x) + [−4A − 4B + 2A] sin(2x) = 0.
So
4A − 2B − 1 = 0,
and
2A + 4B = 0.
So we have
A=
1
,
5
and B = −
1
.
10
So a particular solution to y 00 + 2y 0 + 2y = cos(2x) is:
yp (x) =
1
1
sin(2x) −
cos(2x) .
5
10
VI. When f (x) = m sin(kx) + n cos(kx) is a combination of sines and cosines, and ki is a solution to ar2 + br + c =
c
0(that is, b = 0 and = k 2 ), a particular solution can be like yp (x) = Ax sin(kx)+Bx cos(kx) for some constants
a
A and B.
Example 4. 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
Then the general solution to y 00 + y = 0 is:
y(x) = C1 cos(x) + C2 sin(x).
Let yp (x) = Ax sin(x) + Bx cos(x) be a particular solution to y 00 + y = cos(x), then
yp0 (x)
=
A sin(x) + Ax cos(x) + B cos(x) − Bx sin(x)
yp00 (x)
=
A cos(x) + A cos(x) − Ax sin(x) − B sin(x) − B sin(x) − Bx cos(x)
=
2A cos(x) − Ax sin(x) − 2B sin(x) − Bx cos(x)
=
2A cos(x) − Ax sin(x) − 2B sin(x) − Bx cos(x) + Ax sin(x) + Bx cos(x)
=
2A cos(x) − 2B sin(x)
=
cos(x).
yp00 + yp
LECTURE 14: NONHOMOGENEOUS EQUATIONS
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That is,
[2A − 1] cos(x) − 2B sin(x) = 0.
So
2A − 1 = 0,
and
− 2B = 0.
So we have
A=
1
,
2
and B = 0.
So a particular solution to y 00 + y = cos(x) is:
yp (x) =
x sin(x)
.
2
Remark 1. In summary, if f (x) is one of polynomials, exponential, and a sin(mx) + b cos(mx), then one of g(x), xg(x)
and x2 g(x) can be a special solution, where g(x) is a corresponding one of polynomials (with the same degree as f (x)),
exponential (with the same exponential terms), and A sin(mx) + B cos(mx).
Theorem 2. Let y1,p (x) be a special solution to y 00 + p(x)y + q(x)y = f1 (x), and y2,p (x) be a special solution to
y 00 + p(x)y + q(x)y = f2 (x), then yp (x) = y1,p (x) + y2,p (x) is a special solution to y 00 + p(x)y + q(x)y = f (x), where
f (x) = f1 (x) + f2 (x).
Example 5. Find a special solution to y 00 + y = x + ex .
A special solution to y 00 + y = x can be taken as y(x) = Ax + B, and a special solution to y 00 + y = ex can be
y(x) = Cex , so let yp (x) = Ax + B + Cex be a special solution to y 00 + y = ex , then
yp (x)0
= A + Cex
yp00 (x)
= Cex
yp00 + yp
= Cex + Ax + B + Cex
= Ax + 2Cex + B
= x + ex .
So we get
[2C − 1]ex + [A − 1]x + B = 0.
So
2C − 1 = 0,
2A − 1 = 0,
and B = 0.
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MINGFENG ZHAO
That is,
A = 1,
B = 0,
and C =
1
.
2
Therefore, a special solution to y 00 + y = x + ex can be:
yp (x) = x +
ex
.
2
Variation of Parameters
Definition 1. Let y1 (x) and y2 (x) be two solutions, the Wronskian of y1 and y2 is defined as:
y1 (x) y2 (x) = y1 (x)y20 (x) − y10 (x)y2 (x).
W (y1 , y2 ) = 0
0
y1 (x) y2 (x) Let y1 (x) and y2 (x) be a fundamental set of solutions to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0, consider
the nonhomogeneous equation:
y 00 + p(x)y 0 + q(x)y = f (x).
Now let’s find a particular solution yp (x) to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x): let
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x).
Then
yp0 (x)
= u01 (x)y1 (x) + u02 (x)y2 (x)
+u1 (x)y10 (x) + u2 (x)y20 (x).
Now assume that the sum involving u01 and u02 equals 0, that is, u1 and u2 satisfy
u01 (x)y1 (x) + u02 (x)y2 (x) = 0 .
Then
yp0 (x)
= u1 (x)y10 (x) + u2 (x)y20 (x)
yp00 (x)
= u01 (x)y10 (x) + u1 (x)y100 (x) + u02 (x)y20 (x) + u2 (x)y200 (x)
yp00 (x) + p(x)yp0 (x) + q(x)yp (x)
=
u01 (x)y10 (x) + u1 (x)y100 (x) + u02 (x)y20 (x) + u2 (x)y200 (x)
+p(x) [u1 (x)y10 (x) + u2 (x)y20 (x)] + q(x) [u1 (x)y1 (x) + u2 (x)y2 (x)]
LECTURE 14: NONHOMOGENEOUS EQUATIONS
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= u1 (x) [y100 (x) + p(x)y10 (x) + q(x)y1 (x)]
+u2 (x) [y200 (x) + p(x)y20 (x) + q(x)y2 (x)]
+u01 (x)y10 (x) + u02 (x)y20 (x)
= u01 (x)y10 (x) + u02 (x)y20 (x)
Since y1 (x) and y2 (x) are complementary solutions
= f (x).
So we get
u01 (x)y10 (x) + u02 (x)y20 (x) = f (x).
In summary, u01 and u02 satisfies


 u01 (x)y1 (x) + u02 (x)y2 (x) = 0

 u0 (x)y 0 (x) + u0 (x)y 0 (x) = f (x).
1
1
2
2
Solve the above system, we have


u01 (x)

 =
u02 (x)
=

1
y20 (x)
−y2 (x)
·
−y10 (x)
y1 (x) y2 (x) 0
y1 (x) y20 (x) 

−y2 (x)
f (x)

·
W (y1 , y2 )
y1 (x)
 
·
y1 (x)
0

f (x)
Then we have
−y2 (x)f (x)
,
W (y1 , y2 )
and u02 (x) =
−y2 (x)f (x)
dx,
W (y1 , y2 )
and u2 (x) =
u01 (x) =
y1 (x)f (x)
.
W (y1 , y2 )
Then
Z
u1 (x) =
Z
y1 (x)f (x)
dx.
W (y1 , y2 )
So a particular solution can be:
Z
yp (x) = −y1 (x)
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
Example 6. Find a particular solution of y 00 + y = csc(x).
The characteristic equation of y 00 + y = 0 is:
r2 + 1 = 0.
Z

y1 (x)f (x)
dx .
W (y1 , y2 )
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MINGFENG ZHAO
Solve r2 + 1 = 0, we have
and r2 = −i.
r1 = i,
So the general solution to y 00 + y = 0 is:
y(x) = C1 cos(x) + C2 sin(x).
Let y(x) = u1 (x) cos(x) + u2 (x) sin(x) be particular solution to y 00 + y = csc(x), then
yp0 (x)
= u01 (x) cos(x) − u1 (x) sin(x) + u02 (x) sin(x) + u2 (x) cos(x).
Assume that
u01 (x) cos(x) + u02 (x) sin(x) = 0.
Then
yp0 (x)
= −u1 (x) sin(x) + u2 (x) cos(x)
yp00 (x)
= −u01 (x) sin(x) + u1 (x) cos(x) + u02 (x) cos(x) − u2 (x) sin(x)
yp00 (x) + yp (x)
= −u01 (x) sin(x) + u1 (x) cos(x) + u02 (x) cos(x) − u2 (x) sin(x) + u1 (x) cos(x) + u2 (x) sin(x)
= −u01 (x) sin(x) + u02 (x) cos(x)
=
csc(x).
So u01 and u02 satify


 u01 (x) cos(x) + u02 (x) sin(x) = 0

 −u0 (x) sin(x) + u0 (x) cos(x) = csc(x).
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2
Solve u01 and u02 , we have
u01 (x) = − sin(x) · csc(x) = −1,
and u02 (x) = cos(x) · csc(x) = cot(x).
Then
Z
u1 (x) = −x,
and u2 (x) =
Z
cot(x) dx =
cos(x)
dx = ln sin(x).
sin(x)
Then a particular solution to y 00 + y = csc(x) can be:
yp (x) = −x cos(x) + sin(x) ln sin(x) .
LECTURE 14: NONHOMOGENEOUS EQUATIONS
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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