LECTURE 15: NONHOMOGENEOUS EQUATIONS
MINGFENG ZHAO
October 08, 2014
• 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
– α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the solutions to ar2 + br + c = 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 b2 − 4ac 6= 0, then α = 1.
∗ If m + ki is a root of ar2 + br + c = 0 and b2 − 4ac = 0, then α = 2.
– qn (x) and q̃n (x) are undetermined polynomials with degree n.
Example 1. Find a particular solution to y 00 = x + 1.
The characteristic equation to y 00 = 0 is:
r2 = 0.
Solve r2 = 0, then
r1 = r2 = 0.
Let yp (x) = x2 (Ax + B) = Ax3 + Bx2 be a particular solution to y 00 = x + 1, then
yp0 (x)
=
yp00 (x)
=
3Ax2 + 2Bx
6Ax + 2B
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2
MINGFENG ZHAO
= x + 1.
Then
6A = 1,
and
2B = 1.
That is,
A=
1
,
6
and B =
1
B
So a particular solution to y 00 = x + 1 is:
yp (x) =
x2
x3
+
.
6
2
Example 2. Find a particular solution to y 00 − y = x2 ex .
The characteristic equation of y 00 − y = 0 is:
r2 − 1 = 0.
Then
r1 = −1,
and r2 = 1.
Let yp (x) = x(Ax2 + Bx + C)ex = (Ax3 + Bx2 + Cx)ex be a particular solution to y 00 − y = x2 ex , then
yp0 (x)
=
(3Ax2 + 2Bx + C)ex + (Ax3 + Bx2 + Cx)ex
yp00 (x)
=
(6Ax + 2B)ex + (3Ax2 + 2Bx + C)ex + (3Ax2 + 2Bx + C)ex
+(Ax3 + Bx2 + Cx)ex .
Then
yp00 (x) − yp (x)
=
(6Ax + 2B)ex + (3Ax2 + 2Bx + C)ex + (3Ax2 + 2Bx + C)ex
+(Ax3 + Bx2 + Cx)ex − (Ax3 + Bx2 + Cx)ex
=
(6Ax + 2B)ex + (3Ax2 + 2Bx + C)ex + (3Ax2 + 2Bx + C)ex
=
[6Ax + 2B + 3Ax2 + 2Bx + C + 3Ax2 + 2Bx + C]ex
=
x2 ex
Then
3A + 3A = 1,
6A + 2B + 2B = 0,
and
2B + C + C = 0.
LECTURE 15: NONHOMOGENEOUS EQUATIONS
3
Then
A=
1
,
6
1
B=− ,
4
and C =
1
.
4
Then a particular solution to y 00 − y = x2 ex is:
yp (x) =
x3
x2
x
−
+
6
4
4
ex .
Example 3. Find a particular solution to y 00 − 3y 0 − 4y = −8ex cos(2x).
The characteristic equation of y 00 − 3y 0 − 4y = 0 is:
r2 − 3r − 4 = 0.
Solve r2 − 3r − 4 = 0, we get
r1 = 4,
and r2 = −1.
So 1 + 2i is not a solution to r2 − 3r − 4 = 0. Then let yp (x) = Aex cos(2x) + Bex sin(2x) be a particular
solution to y 00 − 3y 0 − 4y = −8ex cos(2x), then
yp0 (x)
yp00 (x)
= Aex cos(2x) − 2Aex sin(2x) + Bex sin(2x) + 2Bex cos(2x)
=
(A + 2B)ex cos(2x) + [−2A + B]ex sin(2x)
=
(A + 2B)ex cos(2x) − 2(A + 2B)ex sin(2x) + [−2A + B]ex sin(2x) + 2[−2A + B]ex cos(2x)
=
[A + 2B − 4A + 2B]ex cos(2x) + [−2A − 4B − 2A + b]ex sin(2x)
=
[−3A + 4B]ex cos(2x) + [−4A − 3B]ex sin(2x)
Then
yp00 − 3yp0 − 4yp
=
[−3A + 4B]ex cos(2x) + [−4A − 3B]ex sin(2x)
−3 [(A + 2B)ex cos(2x) + [−2A + B]ex sin(2x)]
−4 [Aex cos(2x) + Bex sin(2x)]
=
[−3A + 4B − 3A − 6B − 4A]ex cos(2x) + [−4B − 3B + 6A − 3B − 4B]ex sin(2x)
=
[−10A − 2B]ex cos(2x) + [2A − 10B]ex sin(2x).
So we get
−10A − 2B = −8,
and
2A − 10B = 0.
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MINGFENG ZHAO
Then
A=
10
,
13
and B =
2
.
13
Then a particular solution to y 00 − 3y 0 − 4y = −8ex cos(2x) can be:
yp (x) =
10 x
2
e cos(2x) + ex sin(2x) .
13
13
• Variation of Parameters:
To find a particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x):
I. Find a fundamental set of solutions y1 (x) and y2 (x) to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0.
II. Let yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) be a particular solution to y 00 + p(x)y 0 + q(x)y = f (x)
III. Compute yp0 (x), we get
yp0 (x)
= u01 (x)y1 (x) + u02 (x)y2 (x)
+u1 (x)y10 (x) + u2 (x)y20 (x)
Take
u01 (x)y1 (x) + u2 (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).
IV. Plug yp (x), yp0 (x) and yp00 (x) into y 00 + p(x)y 0 + q(x)y = f (x), we get
u01 (x)y10 (x) + u02 (x)y20 (x) = f (x).
V. Solve u01 (x) and u02 (x) from the system:


 u01 (x)y1 (x) + u02 (x)y2 (x) = 0

 u0 (x)y 0 (x) + u0 (x)y 0 (x) = f (x).
1
2
2
1
Then
u01 (x) =
−y2 (x)f (x)
,
W (y1 , y2 )
and u02 (x) =
y1 (x)f (x)
.
W (y1 , y2 )
LECTURE 15: NONHOMOGENEOUS EQUATIONS
VI. Solve u1 (x) and u2 (x), then
Z
−y2 (x)f (x)
dx,
u1 (x) =
W (y1 , y2 )
Z
and u2 (x) =
y1 (x)f (x)
dx.
W (y1 , y2 )
VII. Write down the solution:
Z
yp (x) = −y1 (x)
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
Z
y1 (x)f (x)
dx .
W (y1 , y2 )
where
y1 (x) y2 (x)
W (y1 , y2 ) = y10 (x) y20 (x)
= y1 (x)y20 (x) − y10 (x)y2 (x).
Example 4. Find a particular solution of y 00 − y = 2 sin(x2 ).
The characteristic equation of y 00 − y = 0 is:
r2 − 1 = 0.
Solve r2 − 1 = 0, then
r1 = 1,
and r2 = −1.
Then the general solution to y 00 − y = 0 is:
y(x) = C1 ex + C2 e−x .
Let yp (x) = u1 (x)ex + u2 (x)e−x be a particular solution to y 00 − y = 2 sin(x2 ), then
yp0 (x) = u01 (x)ex + u1 (x)ex + u02 (x)e−x − u2 (x)e−x
Assume that
u01 (x)ex + u02 (x)e−x = 0.
Then we have
yp0 (x)
= u1 (x)ex − u2 (x)e−x
yp00 (x)
= u01 (x)ex + u1 (x)ex − u02 (x)e−x + u2 (x)e−x
So we have
yp00 − yp
= u01 (x)ex + u1 (x)ex − u02 (x)e−x + u2 (x)e−x − u1 (x)ex + u2 (x)e−x
= u01 (x)ex − u02 (x)e−x
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MINGFENG ZHAO
=
sin(x2 )
In summary, we know that u01 (x) and u02 (x) satisfy


 u01 (x)ex + u02 (x)e−x = 0

 u0 (x)ex − u0 (x)e−x = 2 sin(x2 )
1
2
Solve u01 (x) and u02 (x), we get
u01 (x) = e−x sin(x2 ),
and u02 (x) = −ex sin(x2 ).
So
Z
u1 (x) =
e
−x
2
sin(x ) dx,
Z
and u2 (x) = −
ex sin(x2 ) dx.
Therefore, a particular solution to y 00 − y = 2 sin(x2 ) can be:
Z
Z
yp (x) = ex e−x sin(x2 ) dx − e−x ex sin(x2 ) dx .
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