PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
MINGFENG ZHAO
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.
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)
1
2
MINGFENG ZHAO
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
1
2
2
Then
−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 )
VI. Solve u1 (x) and u2 (x), then
Z
u1 (x) =
Z
y1 (x)f (x)
dx.
W (y1 , y2 )
VII. Write down the solution:
Z
yp (x) = −y1 (x)
where
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
y1 (x) y2 (x)
W (y1 , y2 ) = y10 (x) y20 (x)
Z
y1 (x)f (x)
dx .
W (y1 , y2 )
= y1 (x)y20 (x) − y10 (x)y2 (x).
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
3
Practice Exercises
Exercise 1. Let α be a real constant. a) Find a fundamental set of solutions to y 00 − 4y 0 + α(4 − α)y = 0. (Hint: You
need to consider two cases which depend on α)
Answer: The characteristic equation of y 00 − 4y 0 + α(4 − α) = 0 is:
r2 − 4t + α(4 − α) = 0.
Solve r2 − 4t + α(4 − α) = 0, we get
r1 = α,
and r2 = 4 − α.
We have the following two cases:
I. If r1 = r2 , that is, α = 4 − α, then α = 2. Then y1 (x) = e2x and y2 (x) = xe2x form a fundamental set of
solutions to y 00 − 4y 0 + 4y = 0.
II. If r1 6= r2 , that is, α 6= 4 − α, then α 6= 2. Then y1 (x) = eαx and y2 (x) = e(4−α)x form a fundamental set of
solutions to y 00 − 4y 0 + α(4 − α) = 0.
b) Compute the Wronskian of those two solutions. (Hint: You need to consider two cases which depend on α)
Answer: I. When α = 2, since y1 (x) = e2x and y2 (x) = xe2x , then
2x
e
xe2x
W (y1 , y2 ) = 2e2x e2x + 2xe2x = e2x · e2x + 2xe2x − 2e2x · xe2x
= e4x + 2xe4x − 2xe4x
= e4x .
II. When α 6= 2, since y1 (x) = eαx and y2 (x) = e(4−α)x , then
αx
e
e(4−α)x
W (y1 , y2 ) = αeαx (4 − α)e(4−α)x
= eαx · (4 − α)e(4−α)x − αeαx · e(4−α)x
=
(4 − α)e4x − αe4x
=
(4 − 2α)e4x
4
MINGFENG ZHAO
Exercise 2. a) Find a fundamental set of solutions to y 00 + 10y 0 + 25y = 0.
Answer: The characteristic equation of y 00 + 10y 0 + 25y = 0 is:
r2 + 10r + 25 = 0.
Solve r2 + 10r + 25 = 0, we have
r1 = r2 = −5.
Then y1 (x) = e−5x and y2 (x) = xe−5x form a fundamental set of solutions to y 00 + 10y 0 + 25y = 0.
b) Compute the Wronskian of those two solutions.
Answer: Since y1 (x) = e−5x and y2 (x) = xe−5x , then
e−5x
W (y1 , y2 ) = −5e−5x
xe−5x
e−5x − 5xe−5x
= e−5x · e−5x − 5xe−5x + 5e−5x · xe−5x
= e−10x − 5xe−10x + 5xe−10x
= e−10x .
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
5
Exercise 3. a) Find a fundamental set of solutions to 2y 00 + 4y 0 + 9y = 0.
Answer: The characteristic equation of 2y 00 + 4y 0 + 9y = 0 is:
2r2 + 4r + 9 = 0.
Solve 2r2 + 4r + 9 = 0, we have
√
14
14
r1 =
= −1 +
i, and r2 = −1 −
i.
4
2
2
!
!
√
√
14
14
−x
x and y2 (x) = e sin
x form a fundamental set of solutions to 2y 00 +4y 0 +9y = 0.
2
2
−4 +
Then y1 (x) = e−x cos
√
−56
√
b) Compute the Wronskian of those two
! solutions.
!
√
√
14
14
−x
−x
Answer: Since y1 (x) = e cos
x and y2 (x) = e sin
x , then
2
2
√ √ e−x cos 214 x
e−x sin 214 x
√ √
√ √ √
√ W (y1 , y2 ) = −e−x cos 214 x − 214 e−x sin 214 x
−e−x sin 214 x + 214 e−x cos 214 x
! "
! √
!#
√
√
√
14
14
14 −x
14
−x
−x
= e cos
x · −e sin
x +
e cos
x
2
2
2
2
"
! √
!#
!
√
√
√
14
14 −x
14
14
−x
−x
+ e cos
x +
e sin
x
· e sin
x
2
2
2
2
!
! √
!
√
√
√
14
14
14 −2x
14
2
−2x
x cos
x +
e
cos
x
= −e
sin
2
2
2
2
!
! √
!
√
√
√
14
14
14 −2x 2
14
−2x
+e
sin
x cos
x +
e
sin
x
2
2
2
2
√
14 −2x
e
.
=
2
6
MINGFENG ZHAO
Exercise 4. Find the general solution to 2y 00 + 3y 0 + y = −x + 4x2 .
Answer: The characteristic equation of 2y 00 + 3y 0 + y = 0 is:
2r2 + 3r + 1 = 0.
Solve 2r2 + 3r + 1 = 0, then
r1 = −1,
1
and r2 = − .
2
Then the general solution to 2y 00 + 3y 0 + y = 0 is:
x
y(x) = C1 e−x + C2 e− 2 .
Since 0 is not a solution to 2r2 + 3r + 1 = 0, then we can let yp (x) = Ax2 + Bx + C be a particular solution to
2y 00 + 3y 0 + y = −x + 4x2 , then
yp0 (x)
=
2Ax + B
yp00 (x)
=
2A.
Plug yp , yp0 and yp00 into 2y 00 + 3y 0 + y = −x + 4x2 , then
2yp00 + 3yp0 + yp
=
2 · 2A + 3 · (2Ax + B) + Ax2 + Bx + C
=
Ax2 + (6A + B)x + 4A + 3B + C
=
−x + 4x2 .
Then we have
A = 4,
6A + B = −1,
and
4A + 3B + C = 0.
So we get
A = 4,
B = −25,
and C = 59.
Then
yp (x) = 4x2 − 25x + 59.
Therefore, the general solution to 2y 00 + 3y 0 + y = −x + 4x2 is:
x
y(x) = C1 e−x + C2 e− 2 + 4x2 − 25x + 59.
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
7
Exercise 5. Find the general solution to 2y 00 + y 0 + 2y = 3x2 .
Answer: The characteristic equation of 2y 00 + y 0 + 2y = 0 is:
2r2 + r + 2 = 0.
Solve 2r2 + r + 2 = 0, then
r1 =
−1 +
√
12 − 4 · 2 · 2
1
=− +
4
4
√
15
i,
4
1
and r2 = − −
4
√
15
i.
4
Then the general solution to 2y 00 + y 0 + 2y = 0 is:
√
−x
4
y(x) = C1 e
cos
!
x
15
x + C2 e− 4 sin
4
√
!
15
x .
4
Since 0 is not a solution to 2r2 + r + 2 = 0, then we can let yp (x) = Ax2 + Bx + C be a particular solution to
2y 00 + y 0 + 2y = 3x2 , then
yp0 (x)
=
2Ax + B
yp00 (x)
=
2A.
Plug yp , yp0 and yp00 into 2y 00 + y 0 + 2y = 3x2 , then
2yp00 + yp0 + 2yp
=
2 · 2A + (2Ax + B) + 2(Ax2 + Bx + C)
=
4Ax2 + (2A + 2B)x + 4A + B + 2C
=
3x2 .
Then we have
2A = 3,
2A + 2B = 0,
and
4A + B + 2C = 0.
So we get
A=
3
,
2
3
B=− ,
2
9
and C = − .
4
Then
3 2 3
9
x − x− .
2
2
4
Therefore, the general solution to 2y 00 + 3y 0 + y = −x + 4x2 is:
!
!
√
√
x
15
15
3
3
9
−x
−
y(x) = C1 e 4 cos
x + C2 e 4 sin
x + x2 − x − .
4
4
2
2
4
yp (x) =
8
MINGFENG ZHAO
Exercise 6. Find the general solution to y 00 − 4y 0 + 4y = x.
Answer: The characteristic equation of y 00 − 4y 0 + 4y = 0 is:
r2 − 4r + 4 = 0.
Solve r2 − 4r + 4 = 0, then
r1 = r2 = 2.
Then the general solution to y 00 − 4y 0 + 4y = 0 is:
y(x) = C1 e2x + C2 xe2x .
Since 0 is not a solution to r2 −4r +4 = 0, then we can let yp (x) = Ax+B be a particular solution to y 00 −4y 0 +4y = x,
then
yp0 (x)
= A
yp00 (x)
=
0.
Plug yp , yp0 and yp00 into y 00 − 4y 0 + 4y = x, then
yp00 − 4yp0 + 4yp
=
0 − 4 · A + 4(Ax + B)
=
4Ax − 4A + 4B
= x
Then we have
4A = 1,
− 4A + 4B = 0.
and
So we get
A=
1
,
4
and B =
1
.
4
Then
yp (x) =
x 1
+ .
4 4
Therefore, the general solution to y 00 − 4y 0 + 4y = x is:
y(x) = C1 e2x + C2 xe2x +
x 1
+ .
4 4
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
9
Exercise 7. Find the general solution to y 00 − 3y 0 = 2x + 4.
Answer: The characteristic equation of y 00 − 3y 0 = 0 is:
r2 − 3r = 0.
Solve r2 − 3r = 0, then
r1 = 0,
and r2 = 3.
Then the general solution to y 00 − 3y 0 = 0 is:
y(x) = C1 + C2 e3x .
Since 0 is a solution to r2 − 3r = 0, then we can let yp (x) = x(Ax + B) = Ax2 + Bx be a particular solution to
y 00 − 3y 0 = 2x + 4, then
yp0 (x)
=
2Ax + B
yp00 (x)
=
2A.
Plug yp , yp0 and yp00 into y 00 − 3y 0 = 2x + 4, then
yp00 − 3yp0
=
2A − 3 · (2Ax + B)
=
−6Ax + 2A − 3B
=
2x + 4
Then we have
−6A = 2,
and
2A − 3B = 4.
So we get
1
A=− ,
3
and B = −
14
.
9
Then
yp (x) = −
x2
14
− x.
3
9
Therefore, the general solution to y 00 − 3y 0 = 2x + 4 is:
y(x) = C1 + C2 e3x −
x2
14
− x.
3
9
10
MINGFENG ZHAO
Exercise 8. Find the general solution to 2y 00 = 5x2 + 3.
Answer: The characteristic equation of 2y 00 = 0 is:
2r2 = 0.
Solve 2r2 = 0, then
r1 = r2 = 0.
Then the general solution to 2y 00 = 0 is:
y(x) = C1 + C2 x.
Since 0 is a double solution to 2r2 = 0, then we can let yp (x) = x2 (Ax2 + Bx + C) = Ax4 + Bx3 + Cx2 be a particular
solution to 2y 00 = 5x2 + 3, then
yp0 (x)
=
4Ax3 + 3Bx2 + 2Cx
yp00 (x)
=
12Ax2 + 6Bx + 2C
Plug yp , yp0 and yp00 into 2y 00 = 5x2 + 3, then
2yp00
=
2(12Ax2 + 6Bx + 2C)
=
24Ax2 + 12Bx + 4C
=
5x2 + 3
Then we have
24A = 5,
12B = 0,
and
4C = 3.
So we get
A=
5
,
24
B = 0,
and C =
3
.
4
Then
yp (x) =
5 4 3 2
x + x .
24
4
Therefore, the general solution to 2y 00 = 5x2 + 3 is:
y(x) = C1 + C2 x +
5 4 3 2
x + x .
24
4
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
11
Exercise 9. Find the general solution to y 00 + 2y 0 + 4y = 3e2x .
Answer: The characteristic equation of y 00 + 2y 0 + 4y = 0 is:
r2 + 2r + 4 = 0.
Solve r2 + 2r + 4 = 0, then
r1 =
−2 +
√
√
22 − 4 · 4
= −1 + 3i,
2
and r2 = −1 −
√
3i.
Then the general solution to y 00 + 2y 0 + 4y = 0 is:
√
√
y(x) = C1 e−x cos( 3x) + C2 e−x sin( 3x).
Since 2 is not a solution to r2 +2r+4 = 0, then we can let yp (x) = Ae2x be a particular solution to y 00 +2y 0 +4y = 3e2x ,
then
yp0 (x)
=
2Ae2x
yp00 (x)
=
4Ae2x
Plug yp , yp0 and yp00 into y 00 + 2y 0 + 4y = 3e2x , then
yp00 + 2yp0 + 4yp
=
4Ae2x + 2 · 2Ae2x + 4Ae2x
=
12Ae2x
=
3e2x .
Then we have
12A = 3.
So we get
A=
1
.
4
Then
yp (x) =
1 2x
e .
4
Therefore, the general solution to y 00 + 2y 0 + 4y = 3e2x is:
√
√
1
y(x) = C1 e−x cos( 3x) + C2 e−x sin( 3x) + e2x .
4
12
MINGFENG ZHAO
Exercise 10. Find the general solution to 2y 00 + 4y 0 − 6y = 7ex .
Answer: The characteristic equation of 2y 00 + 4y 0 − 6y = 0 is:
2r2 + 4r − 6 = 0.
Solve 2r2 + 4r − 6 = 0, then
r1 = −3,
and r2 = 1.
Then the general solution to 2y 00 + 4y 0 − 6y = 0 is:
y(x) = C1 e−3x + C2 ex .
Since 1 is a solution to 2r2 + 4r + 6 = 0, then we can let yp (x) = Axex be a particular solution to 2y 00 + 4y 0 − 6y = 7ex ,
then
yp0 (x)
= Aex + Axex
yp00 (x)
= Aex + Aex + Axex
=
2Aex + Axex
Plug yp , yp0 and yp00 into 2y 00 + 4y 0 − 6y = 7ex , then
2yp00 + 4yp0 − 6yp
=
2(2Aex + Axex ) + 4(Aex + Axex ) − 6Axex
=
8Aex
=
7ex .
Then we have
8A = 7.
So we get
A=
7
.
8
Then
yp (x) =
7 x
xe .
8
Therefore, the general solution to 2y 00 + 4y 0 − 6y = 7ex is:
7
y(x) = C1 e−3x + C2 ex + xex .
8
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
13
Exercise 11. Find the general solution to y 00 − 6y 0 + 9y = 5e3x .
Answer: The characteristic equation of y 00 − 6y 0 + 9y = 0 is:
r2 − 6r + 9 = 0 = 0.
Solve r2 − 6r + 9 = 0, then
r1 = r2 = 3.
Then the general solution to y 00 − 6y 0 + 9y = 0 is:
y(x) = C1 e3x + C2 xe3x .
Since 3 is a double solution to r2 − 6r + 9 = 0, then we can let yp (x) = Ax2 e3x be a particular solution to
y 00 − 6y 0 + 9y = 5e3x , then
yp0 (x)
=
2Axe3x + 3Ax2 e3x
yp00 (x)
=
2Ae3x + 6Axe3x + 6Axe3x + 9Ax2 e3x
=
2Ae3x + 12Axe3x + 9Ax2 e3x
Plug yp , yp0 and yp00 into y 00 − 6y 0 + 9y = 5e3x , then
yp00 − 6yp0 + 9yp
=
2Ae3x + 12Axe3x + 9Ax2 e3x − 6(2Axe3x + 3Ax2 e3x ) + 9Ax2 e3x
=
2Ae3x
=
5e3x
Then we have
2A = 5.
So we get
A=
5
.
2
Then
5 2 3x
x e .
2
is:
yp (x) =
Therefore, the general solution to y 00 − 6y 0 + 9y = 5e3x
5
y(x) = C1 e3x + C2 xe3x + x2 e3x .
2
14
MINGFENG ZHAO
Exercise 12. Find the general solution to y 00 + 2y 0 − 5y = 3 sin(2x).
Answer: The characteristic equation of y 00 + 2y 0 − 5y = 0 is:
r2 + 2r − 5 = 0.
Solve r2 + 2r − 5 = 0, then
r1 =
−2 +
√
√
22 + 4 · 5
= −1 + 6,
2
and r2 = −1 −
√
6.
Then the general solution to y 00 + 2y 0 − 5y = 0 is:
y(x) = C1 e(−1+
√
6)x
+ C2 e(−1−
√
6)x
.
Since 2i is not a solution to r2 + 2r − 5 = 0, then we can let yp (x) = A cos(2x) + B sin(2x) be a particular solution
to y 00 + 2y 0 − 5y = 3 sin(2x), then
yp0 (x)
=
−2A sin(2x) + 2B cos(2x)
yp00 (x)
=
−4A cos(2x) − 4B sin(2x).
Plug yp , yp0 and yp00 into y 00 + 2y 0 − 5y = 3 sin(2x), then
yp00 + 2yp0 − 5yp
=
−4A cos(2x) − 4B sin(2x) + 2[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)]
=
[−9A + 4B] cos(2x) + [−4A − 9B] sin(2x)
=
3 sin(2x).
Then we have
−9A + 4B = 0,
and
− 4A − 9B = 3.
So we get
A=−
12
,
97
and B = −
27
.
97
Then
12
27
cos(2x) −
sin(2x).
97
97
Therefore, the general solution to y 00 + 2y 0 − 5y = 3 sin(2x) is:
yp (x) = −
y(x) = C1 e(−1+
√
6)x
+ C2 e(−1−
√
6)x
−
12
27
cos(2x) −
sin(2x).
97
97
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
15
Exercise 13. Find the general solution to 2y 00 + 3y 0 + 10y = sin(4x).
Answer: The characteristic equation of 2y 00 + 3y 0 + 10y = 0 is:
2r2 + 3r + 10 = 0.
Solve 2r2 + 3r + 10 = 0, then
r1 =
−3 +
√
32 − 4 · 2 · 10
3
=− +
4
4
√
71
i,
4
3
and r2 = − −
4
Then the general solution to 2y 00 + 3y 0 + 10y = 0 is:
√
y(x) = C1 e
− 43 x
cos
!
3
71
x + C2 e− 4 x sin
4
√
71
i.
4
√
!
71
x .
4
Since 4i is not a solution to 2r2 + 3r + 10 = 0, then we can let yp (x) = A cos(4x) + B sin(4x) be a particular solution
to 2y 00 + 3y 0 + 10y = sin(4x), then
yp0 (x)
= −4A sin(4x) + 4B cos(4x)
yp00 (x)
= −16A cos(4x) − 16B sin(4x).
Plug yp , yp0 and yp00 into 2y 00 + 3y 0 + 10y = sin(4x), then
2yp00 + 3yp0 + 10yp
=
2[−16A cos(4x) − 16B sin(4x)] + 3[−4A sin(4x) + 4B cos(4x)] + 10[A cos(4x) + B sin(4x)]
=
[−22A + 12B] cos(4x) + [−12A − 22B] sin(4x)
=
sin(4x).
Then we have
−22A + 12B = 0,
and
− 12A − 22B = 1.
So we get
A=−
3
,
157
and B = −
11
.
314
Then
3
11
cos(4x) −
sin(4x).
157
314
Therefore, the general solution to 2y 00 + 3y 0 + 10y = sin(4x) is:
!
!
√
√
71
71
3
11
− 34 x
− 34 x
y(x) = C1 e
cos
x + C2 e
sin
x −
cos(4x) −
sin(4x).
4
4
157
314
yp (x) = −
16
MINGFENG ZHAO
Exercise 14. Find the general solution to y 00 + 9y = 2 sin(3x) + cos(3x).
Answer: The characteristic equation of y 00 + 9y = 0 is:
r2 + 9 = 0.
Solve r2 + 9 = 0, then
r1 = 3i,
and r2 = −3i.
Then the general solution to y 00 + 9y = 0 is:
y(x) = C1 cos(3x) + C2 sin(3x).
Since 3i is a solution to r2 + 9 = 0, then we can let yp (x) = Ax cos(3x) + Bx sin(3x) be a particular solution to
y 00 + 9y = 2 sin(3x) + cos(3x), then
yp0 (x)
=
A cos(3x) − 3Ax sin(3x) + B sin(3x) + 3Bx cos(3x)
yp00 (x)
=
−3A sin(3x) − 3A sin(3x) − 9Ax cos(3x) + 3B cos(3x) + 3B cos(3x) − 9Bx sin(3x)
=
−6A sin(3x) − 9Ax cos(3x) + 6B cos(3x) − 9Bx sin(3x).
Plug yp , yp0 and yp00 into y 00 + 9y = 2 sin(3x) + cos(3x), then
yp00 + 9yp
=
−6A sin(3x) − 9Ax cos(3x) + 6B cos(3x) − 9Bx sin(3x) + 9[Ax cos(3x) + Bx sin(3x)]
=
−6A sin(3x) + 6B cos(3x)
=
2 sin(3x) + cos(3x)
Then we have
−6A = 2,
and
1
A=− ,
3
and B =
6B = 1.
So we get
1
.
6
Then
1
1
yp (x) = − x cos(3x) + x sin(3x).
3
6
00
Therefore, the general solution to y + 9y = 2 sin(3x) + cos(3x) is:
1
1
y(x) = C1 cos(3x) + C2 sin(3x) − x cos(3x) + x sin(3x).
3
6
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
17
Exercise 15. Find the general solution to y 00 + 3y = 2x2 + xe−3x + sin(3x).
Answer: The characteristic equation of y 00 + 3y = 0 is:
r2 + 3 = 0.
Solve r2 + 3 = 0, then
r1 =
√
3i,
√
and r2 = − 3i.
Then the general solution to y 00 + 3y = 0 is:
√
√
y(x) = C1 cos( 3x) + C2 sin( 3x).
Then all of 0, −3 and 3i are not solutions to r2 +3r = 0, let yp (x) = Ax2 +Bx+C+(Dx+E)e−3x +F sin(3x)+G cos(3x)
be a particular solution to y 00 + 3y = 2x2 + xe−3x + sin(3x), then
yp0 (x)
=
2Ax + B + De−3x − 3(Dx + E)e−3x + 3F cos(3x) − 3G sin(3x)
yp00 (x)
=
2A − 3D−3x − 3De−3x + 9(Dx + E)e−3x − 9F sin(3x) − 9G cos(3x)
=
2A − 6De−3x + 9(Dx + E)e−3x − 9F sin(3x) − 9G cos(3x).
Plug yp , yp0 and yp00 into y 00 + 3y = 2x2 + xe−3x + sin(3x), then
yp00 + 3yp
=
2A − 6De−3x + 9(Dx + E)e−3x − 9F sin(3x) − 9G cos(3x)
+3[Ax2 + Bx + C + (Dx + E)e−3x + F sin(3x) + G cos(3x)]
=
3Ax2 + 3Bx + (2A + 3C) + [12Dx − 6D + 12E]e−3x − 6F sin(3x) − 6G cos(3x)
=
2x2 + xe−3x + sin(3x)
Then we have
3A = 2,
3B = 0,
2A + 3C = 0,
12D = 1,
−6D + 12E = 0,
−6F = 1,
and
− 6G = 0.
So we get
A=
2
,
3
B = 0,
4
C=− ,
9
D=
1
,
12
E=
1
,
24
1
F =− ,
6
and G = 0.
Then
2 2 4
1
1
1
x − + xe−3x + e−3x − sin(3x).
3
9 12
24
6
Therefore, the general solution to y 00 + 3y = 2x2 + xe−3x + sin(3x) is:
yp (x) =
√
√
2
4
1
1
1
y(x) = C1 cos( 3x) + C2 sin( 3x) + x2 − + xe−3x + e−3x − sin(3x).
3
9 12
24
6
18
MINGFENG ZHAO
Exercise 16. Find the general solution to y 00 + 4y = 4 sin(2x) + 3.
Answer: The characteristic equation of y 00 + 4y = 0 is:
r2 + 4 = 0.
Solve r2 + 4 = 0, then
r1 = 2i,
and r2 = −2i.
Then the general solution to y 00 + 4y = 0 is:
y(x) = C1 cos(2x) + C2 sin(2x).
Then 2i is a solution to r2 + 4 = 0, but 0 is not a solution to r2 + 4 = 0. Let yp (x) = Ax cos(2x) + Bx sin(2x) + C
be a particular solution to y 00 + 4y = 4 sin(2x) + 3, then
yp0 (x)
=
A cos(2x) − 2Ax sin(2x) + B sin(2x) + 2Bx cos(2x)
yp00 (x)
=
−2A sin(2x) − 2A sin(2x) − 4Ax cos(2x) + 2B cos(2x) + 2B cos(2x) − 4Bx sin(2x)
=
−4A sin(2x) − 4Ax cos(2x) + 4B cos(2x) − 4Bx sin(2x).
Plug yp , yp0 and yp00 into y 00 + 4y = 4 sin(2x) + 3, then
yp00 + 4yp
=
−4A sin(2x) − 4Ax cos(2x) + 4B cos(2x) − 4Bx sin(2x) + 4[Ax cos(2x) + Bx sin(2x) + C]
=
−4A sin(2x) + 4B cos(2x) + 4C
=
4 sin(2x) + 3
Then we have
−4A = 4,
4B = 0,
and
A = −1,
B = 0,
and C =
4C = 3.
So we get
3
4
Then
3
yp (x) = −x cos(2x) + .
4
00
Therefore, the general solution to y + 4y = 4 sin(2x) + 3 is:
3
y(x) = C1 cos(2x) + C2 sin(2x) − x cos(2x) + .
4
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
Exercise 17. Solve the following problem:
y 00 + y = f (x),
y(0) = 0,
y 0 (0) = 1,
where

 x,
if 0 ≤ x ≤ π,
f (x) =
 πeπ−x , if x > π.
Answer: The characteristic equation of y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, then
and r2 = −i.
r1 = i,
Then the general solution to y 00 + y = 0 is:
y(x) = C1 cos(x) + C2 sin(x).
Then 0 is a not solution to r2 + 1 = 0. Let yp (x) = Ax + B be a particular solution to y 00 + y = x, then
yp0 (x)
=
A
yp00 (x) = 0.
Plug yp , yp0 and yp00 into y 00 + y = x, then
yp00 + yp = Ax + B = x.
So
A = 1,
and B = 0.
Then
yp (x) = x.
So the general solution to y 00 + y = x is:
y(x) = C1 cos(x) + C2 sin(x) + x.
For 0 ≤ x ≤ π, then
y 0 (x) = −C1 sin(x) + C2 cos(x) + 1.
Since y(0) = 0 and y 0 (0) = 1, then
C1 = 0,
and C2 + 1 = 1.
19
20
MINGFENG ZHAO
Then C1 = 0 and C2 = 0, that is,
∀0 ≤ x ≤ π.
y(x) = x,
So y 0 (x) = 1, which implies that
y(π) = π,
and y 0 (π) = 1.
Since −1 is not a solution to r2 + 1 = 0, then let yp (x) = Ae−x be a particular solution to y 00 + y = πeπ−x . So we get
yp0 (x)
=
−Ae−x
yp00 (x)
=
Ae−x .
Plug yp , yp0 and yp00 into y 00 + y = πeπ−x , then
yp00 + yp = Ae−x + Ae−x = 2Ae−x = πeπ−x .
Then
A=
πeπ
.
2
So
yp (x) =
πeπ −x
e .
2
Then the general solution to y 00 + y = πeπ−x is:
y(x) = C1 cos(x) + C2 sin(x) +
πeπ −x
e .
2
So for x ≥ π, we have
y 0 (x) = −C1 sin(x) + C2 cos(x) −
πeπ −x
e .
2
Since y(π) = π and y 0 (π) = 1, then
−C1 +
π
= π,
2
and
− C2 −
π
= 1.
2
So we get
π
C1 = − ,
2
and C2 = −1 −
π
.
2
That is,
π
2+π
πeπ −x
cos(x) −
sin(x) +
e , ∀x ≥ π.
2
2
2
In summary, the solution to y 00 + y = f (x), y(0) = 0 and y 0 (0) = 1 is:

 x,
if 0 ≤ x ≤ π,
π
y(x) =
π
2
+
π
πe
−x
 − cos(x) −
sin(x) +
e , if x > π.
2
2
2
y(x) = −
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
21
Exercise 18. Find the general solution to y 00 + 4y = (6x + 7) cos(2x).
Answer: The characteristic equation of y 00 + 4y = 0 is:
r2 + 4 = 0.
Solve r2 + 4 = 0, then
r1 = 2i,
and r2 = −2i.
Then the general solution to y 00 + 4y = 0 is:
y(x) = C1 cos(2x) + C2 sin(2x).
Since 2i is a solution to r2 + 4 = 0, then we can let yp (x) = x(Ax + B) cos(2x) + x(Cx + D) sin(2x) = (Ax2 +
Bx) cos(2x) + (Cx2 + Dx) sin(2x) be a particular solution to y 00 + 4y = (6x + 7) cos(2x), then
yp0 (x)
=
(2Ax + B) cos(2x) − 2(Ax2 + Bx) sin(2x) + (2Cx + D) sin(2x) + 2(Cx2 + Dx) cos(2x)
yp00 (x)
=
2A cos(2x) − 2(2Ax + B) sin(2x) − 2(2Ax + B) sin(2x) − 4(Ax2 + Bx) cos(2x)
+2C sin(2x) + 2(2Cx + D) cos(2x) + 2(2Cx + D) cos(2x) − 4(Cx2 + Dx) sin(2x)
[−4Ax2 + (−4B + 8C)x + 2A + 4D] cos(2x) + [−4Cx2 + (−8A − 4D)x − 4B + 2C] sin(2x).
=
Plug yp , yp0 and yp00 into y 00 + 4y = (6x + 7) cos(2x), then
yp00 + 4yp
=
[−4Ax2 + (−4B + 8C)x + 2A + 4D] cos(2x) + [−4Cx2 + (−8A − 4D)x − 4B + 2C] sin(2x)
+4[(Ax2 + Bx) cos(2x) + (Cx2 + Dx) sin(2x)]
=
[8Cx + 2A + 4D] cos(2x) + [−8Ax − 4B + 2C] sin(2x)
=
(6x + 7) cos(2x).
Then we have
8C = 6,
−8A = 0,
2A + 4D = 7,
and
− 4B + 2C = 0.
So we get
A = 0,
B=
3
,
8
C=
3
4
and D =
7
.
4
Then
3
3
7
x cos(2x) + x2 sin(2x) + x sin(2x).
8
4
4
Therefore, the general solution to y 00 + 4y = (6x + 7) cos(2x) is:
yp (x) =
3
3
7
y(x) = C1 cos(2x) + C2 sin(2x) + x cos(2x) + x2 sin(2x) + x sin(2x).
8
4
4
22
MINGFENG ZHAO
Exercise 19. Find the general solution to y 00 + 2y 0 + 5y = 3xe−x cos(2x).
Answer: The characteristic equation of y 00 + 2y 0 + 5y = 0 is:
r2 + 2r + 5 = 0.
Solve r2 + 2r + 5 = 0, then
r1 =
−2 +
√
22 − 4 · 5
= −1 + 2i,
2
and r2 = −1 − 2i.
Then the general solution to y 00 + 2y 0 + 5y = 0 is:
y(x) = C1 e−x cos(2x) + C2 e−x sin(2x).
Since −1 + 2i is a solution to r2 + 2r + 5 = 0, then we can let yp (x) = x(Ax + B)e−x cos(2x) + x(Cx + D)e−x sin(2x) =
(Ax2 + Bx)e−x cos(2x) + (Cx2 + Dx)e−x sin(2x) be a particular solution to y 00 + 2y 0 + 5y = 3xe−x cos(2x), then
yp0 (x)
=
(2Ax + B)e−x cos(2x) − (Ax2 + Bx)e−x cos(2x) − 2(Ax2 + Bx)e−x sin(2x)
+(2Cx + D)e−x sin(2x) − (Cx2 + Dx)e−x sin(2x) + 2(Cx2 + Dx)e−x cos(2x)
=
[(−A + 2C)x2 + (2A − B + 2D)x + B]e−x cos(2x)
+[(−2A − C)x2 + (−2B + 2C − D)x + D]e−x sin(2x)
yp00 (x)
=
[2(−A + 2C)x + (2A − B + 2D)]e−x cos(2x) − [(−A + 2C)x2 + (2A − B + 2D)x + B]e−x cos(2x)
−2[(−A + 2C)x2 + (2A − B + 2D)x + B]e−x sin(2x) + [2(−2A − C)x + (−2B + 2C − D)]e−x sin(2x)
−[(−2A − C)x2 + (−2B + 2C − D)x + D]e−x sin(2x) + 2[(−2A − C)x2 + (−2B + 2C − D)x + D]e−x cos(2x)
=
[(−3A − 4C)x2 + (−4A − 3B + 8C − 4D)x + 2A − 2B + 4D]e−x cos(2x)
+[(4A − 3C)x2 + (−8A + 4B − 4C + 5D)x − 4B + 2C − 2D]e−x sin(2x).
Plug yp , yp0 and yp00 into y 00 + 2y 0 + 5y = 3xe−x cos(2x), then
yp00 + 2yp0 + 5yp
=
[(−3A − 4C)x2 + (−4A − 3B + 8C − 4D)x + 2A − 2B + 4D]e−x cos(2x)
+[(4A − 3C)x2 + (−8A + 4B − 4C + 5D)x − 4B + 2C − 2D]e−x sin(2x)
+2[(−A + 2C)x2 + (2A − B + 2D)x + B]e−x cos(2x)
+2[(−2A − C)x2 + (−2B + 2C − D)x + D]e−x sin(2x)
+5(Ax2 + Bx)e−x cos(2x) + 5(Cx2 + Dx)e−x sin(2x)
=
[8Cx + 2A + 4D]e−x cos(2x) + [(−8A + 8D)x − 4B + 2C]e−x sin(2x)
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
Then we have
8C = 3,
2A + 4D = 0,
−8A + 8D = 0,
and
− 4B + 2C = 0.
So we get
A = 0,
B=
3
,
16
C=
3
8
and D = 0.
Then
3 −x
3
xe cos(2x) + x2 sin(2x).
16
8
00
0
−x
Therefore, the general solution to y + 2y + 5y = 3xe cos(2x) is:
yp (x) =
y(x) = C1 e−x cos(2x) + C2 e−x sin(2x) +
3 −x
3
xe cos(2x) + x2 sin(2x).
16
8
23
24
MINGFENG ZHAO
Exercise 20. Find the general solution to y 00 + 4y 0 + 4y = x−2 e−2x .
Answer: The characteristic equation of y 00 + 4y 0 + 4y = 0 is:
r2 + 4r + 4 = 0.
Solve r2 + 4r + 4 = 0, then
r1 = r2 = −2.
Then the general solution to y 00 + 4y 0 + 4y = 0 is:
y(x) = C1 e−2x + C2 xe−2x .
Then y1 (x) = e−2x and y20 (x) = xe−2x form a fundamental set of solutions to y 00 + 4y 0 + 4y = 0. Let yp (x) =
u1 (x)y1 (x) + u2 (x)y2 (x) = u1 (x)e−2x + u2 (x)xe−2x be a particular solution to y 00 + 4y 0 + 4y = x−2 e−2x such that
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x)e−2x + u02 (x)xe−2x = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (−2e−2x ) + u02 (x) · (e−2x − 2xe−2x ) = x−2 e−2x .
So we get

 u0 (x) + xu0 (x) = 0
1
2
 −2u0 (x) + (1 − 2x)u0 (x) = x−2 .
1
2
Solve u01 (x) and u02 (x), we get
u01 (x) = −x−1 ,
and u02 (x) = x−2 .
x−1 dx = − ln |x|,
and u02 (x) =
Then
Z
u1 (x) = −
Z
1
x−2 dx = − .
x
So
yp (x) = −e−2x ln |x| − e−2x .
Therefore, the general solution to y 00 + 4y 0 + 4y = x−2 e−2x is:
y(x) = C1 e−2x + C2 xe−2x − e−2x ln |x|.
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
Exercise 21. Find the general solution to y 00 − 2y 0 + y =
25
ex
.
1 + x2
Answer: The characteristic equation of y 00 − 2y 0 + y = 0 is:
r2 − 2r + 1 = 0.
Solve r2 − 2r + 1 = 0, then
r1 = r2 = 1.
Then the general solution to y 00 − 2y 0 + y = 0 is:
y(x) = C1 ex + C2 xex .
Then y1 (x) = ex and y20 (x) = xex form a fundamental set of solutions to y 00 − 2y 0 + y = 0. Let yp (x) = u1 (x)y1 (x) +
ex
u2 (x)y2 (x) = u1 (x)ex + u2 (x)xex be a particular solution to y 00 − 2y 0 + y =
such that
1 + x2
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x)ex + u02 (x)xex = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (ex ) + u02 (x) · (ex + xex ) =
ex
.
1 + x2
So we get

 u01 (x) + xu02 (x) = 0
 u0 (x) + (1 + x)u0 (x) =
1
2
1
.
1 + x2
Solve u01 (x) and u02 (x), we get
u01 (x) = −
x2
x
,
+1
and u02 (x) =
x2
1
.
+1
Then
Z
u1 (x) = −
1
x
dx = ln(x2 + 1),
x2 + 1
2
and
u02 (x)
Z
=
1
dx = arctan(x).
x2 + 1
So
ex
ln(x2 + 1) + xex arctan(x).
2
ex
Therefore, the general solution to y 00 − 2y 0 + y =
is:
1 + x2
ex
y(x) = C1 ex + C2 xex +
ln(x2 + 1) + xex arctan(x).
2
yp (x) =
26
MINGFENG ZHAO
Exercise 22. Find the general solution to y 00 + 4y = 3 csc(2x).
Answer: The characteristic equation of y 00 + 4y = 0 is:
r2 + 4 = 0.
Solve r2 + 4 = 0, then
and r2 = −2i.
r1 = 2i,
Then the general solution to y 00 + 4y = 0 is:
y(x) = C1 cos(2x) + C2 sin(2x).
Then y1 (x) = cos(2x) and y20 (x) = sin(2x) form a fundamental set of solutions to y 00 + 4y = 0. Let yp (x) =
u1 (x)y1 (x) + u2 (x)y2 (x) = u1 (x) cos(2x) + u2 (x) sin(2x) be a particular solution to y 00 + 4y = 3 csc(2x) such that
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x) cos(2x) + u02 (x) sin(2x) = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (−2 sin(2x)) + u02 (x) · (2 cos(2x)) = 3 csc(2x).
So we get

 u0 (x) cos(2x) + u0 (x) sin(2x) = 0
1
2
 −2 sin(2x)u0 (x) + 2 cos(2x)u0 (x) = 3 csc(2x).
1
2
Solve u01 (x) and u02 (x), we get
u01 (x) = −3 cos(2x)
and u02 (x) =
3 cos2 (2x)
.
sin(2x)
Then
Z
u1 (x) = −
3
3 cos(2x) dx = − sin(2x),
2
and u02 (x) =
Z
3 cos2 (2x)
dx = 3
sin(2x)
Z
cos2 (2x)
dx.
sin(2x)
So
3
yp (x) = − sin(2x) cos(2x) + 3 sin(2x)
2
Z
cos2 (2x)
dx.
sin(2x)
Therefore, the general solution to y 00 + 4y = 3 csc(2x) is:
y(x) = C1 cos(2x) + C2 sin(2x) −
3
sin(2x) cos(2x) + 3 sin(2x)
2
Z
cos2 (2x)
dx.
sin(2x)
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
27
Exercise 23. Find the general solution to y 00 + y = tan(x).
Answer: The characteristic equation of y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, then
r1 = i,
and r2 = −i.
Then the general solution to y 00 + y = 0 is:
y(x) = C1 cos(x) + C2 sin(x).
Then y1 (x) = cos(x) and y20 (x) = sin(x) form a fundamental set of solutions to y 00 + y = 0. Let yp (x) = u1 (x)y1 (x) +
u2 (x)y2 (x) = u1 (x) cos(x) + u2 (x) sin(x) be a particular solution to y 00 + y = tan(x) such that
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x) cos(x) + u02 (x) sin(x) = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (− sin(x)) + u02 (x) · (cos(x)) = tan(x).
So we get

 u0 (x) cos(x) + u0 (x) sin(x) = 0
1
2
 − sin(x)u0 (x) + cos(x)u0 (x) = tan(x).
1
Solve u01 (x) and u02 (x), we get
u01 (x) = −
2
sin2 (x)
cos(x)
and u02 (x) = sin(x).
Then
Z
u1 (x) = −
sin2 (x)
dx,
cos(x)
and u02 (x) =
Z
sin(x) dx = − cos(x).
So
Z
yp (x) = − cos(x)
sin2 (x)
dx − sin(x) cos(x).
cos(x)
Therefore, the general solution to y 00 + y = tan(x) is:
Z
y(x) = C1 cos(x) + C2 sin(x) − cos(x)
sin2 (x)
dx − sin(x) cos(x).
cos(x)
28
MINGFENG ZHAO
Exercise 24. Solve the following problem:
y 00 − 2y 0 + y = f (x),
where
y(0) = 0,
y 0 (0) = 0,

 xex ,
if 0 ≤ x ≤ 1,
f (x) =
 esin( πx
)
2
, if x > 1.
Answer: The characteristic equation of y 00 − 2y 0 + y = 0 is:
r2 − 2r + 1 = 0.
Solve r2 − 2r + 1 = 0, then
r1 = r2 = 1.
Then the general solution to y 00 − 2y 0 + y = 0 is:
y(x) = C1 ex + C2 xex .
Then y1 (x) = ex and y20 (x) = xex form a fundamental set of solutions to y 00 − 2y 0 + y = 0. Since 1 is a double zero
of r2 − 2r + 1 = 0, let yp (x) = x2 (Ax + B)ex = (Ax3 + Bx2 )ex be a particular solution to y 00 − 2y 0 + y = xex , then
yp0 (x)
yp00 (x)
=
(3Ax2 + 2Bx)ex + (Ax3 + Bx2 )ex
=
[Ax3 + (3A + B)x2 + 2Bx]ex
=
(6Ax + 2B)ex + (3Ax2 + 2Bx)ex + (3Ax2 + 2Bx)ex + (Ax3 + Bx2 )ex
=
[Ax3 + (6A + B)x2 + (6A + 4B)x + 2B]ex .
Plug yp , yp0 and yp00 into y 00 − 2y 0 + y = xex , then
yp00 − 2yp0 + yp
=
[Ax3 + (6A + B)x2 + (6A + 4B)x + 2B]ex − 2[Ax3 + (3A + B)x2 + 2Bx]ex + (Ax3 + Bx2 )ex
=
[6Ax + 2B]ex
= xex .
Then we have
6A = 1,
and
2B = 0.
So we get
A=
1
,
6
and B = 0.
Then
yp (x) =
1 3 x
x e .
6
PRACTICE EXERCISES FOR NONHOMOGENEOUS EQUATIONS
29
So the general solution to y 00 − 2y 0 + y = xex is:
1
y(x) = C1 ex + C2 xex + x3 ex .
6
Then
1
1
y 0 (x) = C1 ex + C2 ex + C2 xex + x2 ex + x3 ex .
2
6
Since y(0) = 0 and y 0 (0) = 0, then
C1 = 0,
and C1 + C2 = 0.
That is, C1 = C2 = 0, which implies that
y(x) =
1 3 x
x e ,
6
y(1) =
e
,
6
∀0 ≤ x ≤ 1.
So we get
and y 0 (1) =
2e
.
3
Since y1 (x) = ex and y20 (x) = xex form a fundamental set of solutions to y 00 − 2y 0 + y = 0, let yp (x) = u1 (x)y1 (x) +
πx
u2 (x)y2 (x) = u1 (x)ex + u2 (x)xex be a particular solution to y 00 − 2y 0 + y = esin( 2 ) such that
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x)ex + u02 (x)xex = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (ex ) + u02 (x) · (ex + xex ) = esin(
πx
2
).
So we get

 u0 (x) + xu0 (x) = 0
1
2
 u0 (x) + (1 + x)u0 (x) = esin( πx
2 )−x .
1
2
Solve u01 (x) and u02 (x), we get
u01 (x) = −xesin(
πx
2
)−x ,
and u02 (x) = esin(
πx
2
)−x .
Then
Z
u1 (x) = −
xesin(
πx
2
)−x dx,
and u02 (x) =
Z
esin(
πx
2
)−x dx.
So
yp (x) = −ex
Z
x
πt
tesin( 2 )−t dt + xex
1
Z
x
πt
esin( 2 )−t dt.
1
Therefore, the general solution to y 00 − 2y 0 + y = esin( 2 ) is:
Z x
Z
sin( πt
−t
x
x
x
x
)
2
y(x) = C1 e + C2 xe − e
te
dt + xe
πx
1
1
x
πt
esin( 2 )−t dt.
30
MINGFENG ZHAO
Then
0
y (x)
x
x
x
= C1 e + C2 e + C2 xe − e
x
x
Z
πt
πx
tesin( 2 )−t dt − ex xesin( 2 )−x
1
+ex
Z
x
esin( )−t dt + xex
πt
2
1
Since y(1) =
Z
x
πt
πx
esin( 2 )−t dt + xex esin( 2 )−x
1
2e
e
and y 0 (1) = , then
6
3
e
,
6
C1 e + C2 e =
and C1 e + C2 e + C2 e =
2e
.
3
So we have
1
C1 = − ,
3
and C2 =
1
2
That is, we have
1
1
y(x) = − ex + xex − ex
3
2
Z
1
x
πt
tesin( 2 )−t dt + xex
Z
x
πt
esin( 2 )−t dt,
∀x ≥ 1.
1
In summary, the solution to y 00 − 2y 0 + y = f (x), y(0) = 0 and y 0 (0) = 0 is:

1

 x3 ex ,
if 0 ≤ x ≤ 1,
6
Z x
Z x
y(x) =
πt
πt
1
1

 − ex + xex − ex
esin( 2 )−t dt, if x ≥ 1.
tesin( 2 )−t dt + xex
3
2
1
1
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