Second Order Nonhomogeneous Linear
Differential Equations with Constant
Coefficients:
the method of undetermined coefficients
Xu-Yan Chen
◮ Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).
◮ General solution structure:
y(t) = yp(t) + yc(t)
where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:
a2y′′
c (t) + a1y′
c(t) + a0yc(t) = 0.
◮ The characteristic roots: a2λ2 + a1λ + a0 = 0
⇒ The complementary solutions yc(t).
◮ Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).
◮ General solution structure:
y(t) = yp(t) + yc(t)
where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:
a2y′′
c (t) + a1y′
c(t) + a0yc(t) = 0.
◮ The characteristic roots: a2λ2 + a1λ + a0 = 0
⇒ The complementary solutions yc(t).
◮ Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).
◮ General solution structure:
y(t) = yp(t) + yc(t)
where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:
a2y′′
c (t) + a1y′
c(t) + a0yc(t) = 0.
◮ The characteristic roots: a2λ2 + a1λ + a0 = 0
⇒ The complementary solutions yc(t).
◮ What is this note about? The Method ofUndetermined Coefficients:
◮ Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).
◮ General solution structure:
y(t) = yp(t) + yc(t)
where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:
a2y′′
c (t) + a1y′
c(t) + a0yc(t) = 0.
◮ The characteristic roots: a2λ2 + a1λ + a0 = 0
⇒ The complementary solutions yc(t).
◮ What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.
◮ Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).
◮ General solution structure:
y(t) = yp(t) + yc(t)
where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:
a2y′′
c (t) + a1y′
c(t) + a0yc(t) = 0.
◮ The characteristic roots: a2λ2 + a1λ + a0 = 0
⇒ The complementary solutions yc(t).
◮ What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.
◮ Main Idea: Set up a trial function yp(t), by copying thefunction form of f(t).
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
3y′′
c + y′
c − 2yc = 0 (the corresponding homog eq)
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
3y′′
c + y′
c − 2yc = 0 (the corresponding homog eq)
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = ae4t (form copied from f(t) = 10e4t)
where a is the undetermined coefficient.
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
3y′′
c + y′
c − 2yc = 0 (the corresponding homog eq)
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = ae4t (form copied from f(t) = 10e4t)
where a is the undetermined coefficient.
◮ Substitute yp(t) in the nonhomog eq:
3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t
= 3(16ae4t) + (4ae4t) − 2ae4t
= 50ae4t
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
3y′′
c + y′
c − 2yc = 0 (the corresponding homog eq)
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = ae4t (form copied from f(t) = 10e4t)
where a is the undetermined coefficient.
◮ Substitute yp(t) in the nonhomog eq:
3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t
= 3(16ae4t) + (4ae4t) − 2ae4t
= 50ae4t
◮ Compare the coefficients of the two sides:
50a = 10 ⇒ a =1
5
Example 1: Solve 3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ General solutions y(t) = yc(t) + yp(t).
◮ Find complementary solutions yc(t):
3y′′
c + y′
c − 2yc = 0 (the corresponding homog eq)
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = ae4t (form copied from f(t) = 10e4t)
where a is the undetermined coefficient.
◮ Substitute yp(t) in the nonhomog eq:
3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t
= 3(16ae4t) + (4ae4t) − 2ae4t
= 50ae4t
◮ Compare the coefficients of the two sides:
50a = 10 ⇒ a =1
5⇒ yp(t) =
1
5e4t
Example 1 (continued): Solve3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ Combine yc and yp to get
Gen Sols of Nonhomg Eq: y(t) =1
5e4t + C1e
−t + C2e2
3t.
Example 1 (continued): Solve3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ Combine yc and yp to get
Gen Sols of Nonhomg Eq: y(t) =1
5e4t + C1e
−t + C2e2
3t.
◮ Use initial conditions:
y(0) = −1 ⇒ 1
5+ C1 + C2 = −1
y′(t) = 4
5e4t − C1e
−t + 2
3C2e
2
3t, y′(0) = 3 ⇒ 4
5− C1 + 2
3C2 = 3
Example 1 (continued): Solve3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ Combine yc and yp to get
Gen Sols of Nonhomg Eq: y(t) =1
5e4t + C1e
−t + C2e2
3t.
◮ Use initial conditions:
y(0) = −1 ⇒ 1
5+ C1 + C2 = −1
y′(t) = 4
5e4t − C1e
−t + 2
3C2e
2
3t, y′(0) = 3 ⇒ 4
5− C1 + 2
3C2 = 3
Solve this:
{
C1 = − 9
5
C2 = 3
5
Example 1 (continued): Solve3y′′ + y′ − 2y = 10e4t, y(0) = −1, y′(0) = 3.
◮ Combine yc and yp to get
Gen Sols of Nonhomg Eq: y(t) =1
5e4t + C1e
−t + C2e2
3t.
◮ Use initial conditions:
y(0) = −1 ⇒ 1
5+ C1 + C2 = −1
y′(t) = 4
5e4t − C1e
−t + 2
3C2e
2
3t, y′(0) = 3 ⇒ 4
5− C1 + 2
3C2 = 3
Solve this:
{
C1 = − 9
5
C2 = 3
5
◮ The solution of the initial value problem:
y(t) =1
5e4t −
9
5e−t +
3
5e
2
3t.
Nonhomogeneous Linear Equations:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
Towards the Rules of Setting Up the Trial Function:
f(t) yp(t)
kert Aert
· · · · · ·
(to be continued)
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = Ate−2t.
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = Ate−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3(Ate−2t)′′ + (Ate−2t)′ − 2Ate−2t
= 3(−4Ae−2t + 4Ate−2t) + (Ae−2t − 2Ate−2t) − 2Ate−2t
= −11Ae−2t + 8Ate−2t
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = Ate−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3(Ate−2t)′′ + (Ate−2t)′ − 2Ate−2t
= 3(−4Ae−2t + 4Ate−2t) + (Ae−2t − 2Ate−2t) − 2Ate−2t
= −11Ae−2t + 8Ate−2t
◮ Compare the coefficients of the two sides:{
−11A = 08A = −8
⇒ Impossible!
Example 2: Solve 3y′′ + y′ − 2y = −8te−2t.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = Ate−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3(Ate−2t)′′ + (Ate−2t)′ − 2Ate−2t
= 3(−4Ae−2t + 4Ate−2t) + (Ae−2t − 2Ate−2t) − 2Ate−2t
= −11Ae−2t + 8Ate−2t
◮ Compare the coefficients of the two sides:{
−11A = 08A = −8
⇒ Impossible!
The choice of the trial function yp(t) = Ate−2t was WRONG!
Example 2 (continued): Solve 3y′′ + y′ − 2y = −8te−2t.
◮ The correct point of view:
f(t) = −8te−2t = (a polynomial of degree one)e−2t.
Example 2 (continued): Solve 3y′′ + y′ − 2y = −8te−2t.
◮ The correct point of view:
f(t) = −8te−2t = (a polynomial of degree one)e−2t.
◮ The correct trial function:yp(t) = (A + Bt)e−2t.
Example 2 (continued): Solve 3y′′ + y′ − 2y = −8te−2t.
◮ The correct point of view:
f(t) = −8te−2t = (a polynomial of degree one)e−2t.
◮ The correct trial function:yp(t) = (A + Bt)e−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3[(A + Bt)e−2t]′′ + [(A + Bt)e−2t]′ − 2(A + Bt)e−2t
= 3(4A − 4B + 4Bt)e−2t + (−2A + B − 2Bt)e−2t
+(−2A − 2Bt)e−2t
= (8A − 11B)e−2t + 8Bte−2t
Example 2 (continued): Solve 3y′′ + y′ − 2y = −8te−2t.
◮ The correct point of view:
f(t) = −8te−2t = (a polynomial of degree one)e−2t.
◮ The correct trial function:yp(t) = (A + Bt)e−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3[(A + Bt)e−2t]′′ + [(A + Bt)e−2t]′ − 2(A + Bt)e−2t
= 3(4A − 4B + 4Bt)e−2t + (−2A + B − 2Bt)e−2t
+(−2A − 2Bt)e−2t
= (8A − 11B)e−2t + 8Bte−2t
◮ Compare the coefficients of the two sides:{
8A − 11B = 0
8B = −8⇒
{
A = − 11
8
B = −1⇒ yp(t) =
(
− 11
8− t
)
e−2t
Example 2 (continued): Solve 3y′′ + y′ − 2y = −8te−2t.
◮ The correct point of view:
f(t) = −8te−2t = (a polynomial of degree one)e−2t.
◮ The correct trial function:yp(t) = (A + Bt)e−2t.
◮ Substitute yp(t) in the nonhomog eq:
−8te−2t = 3[(A + Bt)e−2t]′′ + [(A + Bt)e−2t]′ − 2(A + Bt)e−2t
= 3(4A − 4B + 4Bt)e−2t + (−2A + B − 2Bt)e−2t
+(−2A − 2Bt)e−2t
= (8A − 11B)e−2t + 8Bte−2t
◮ Compare the coefficients of the two sides:{
8A − 11B = 0
8B = −8⇒
{
A = − 11
8
B = −1⇒ yp(t) =
(
− 11
8− t
)
e−2t
◮ The General Solutions of the Nonhomogeneous Equation:
y(t) = yp(t) + yc(t) =(
− 11
8− t
)
e−2t + C1e−t + C2e
2
3t.
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2.
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2. This does not work!
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2. This does not work!
◮ The correct trial function:yp(t) = A + Bt + Ct2.
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2. This does not work!
◮ The correct trial function:yp(t) = A + Bt + Ct2.
◮ Substitute yp(t) in the nonhomog eq:
−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)
= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2. This does not work!
◮ The correct trial function:yp(t) = A + Bt + Ct2.
◮ Substitute yp(t) in the nonhomog eq:
−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)
= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2
◮ Compare the coefficients of the two sides:
−2A + B + 6C = 0
−2B + 2C = 0
−2C = −12
⇒
A = 21
B = 6
C = 6
⇒ yp(t) = 21 + 6t + 6t2
Example 3: Solve 3y′′ + y′ − 2y = −12t2.
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial functionyp(t) = At2. This does not work!
◮ The correct trial function:yp(t) = A + Bt + Ct2.
◮ Substitute yp(t) in the nonhomog eq:
−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)
= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2
◮ Compare the coefficients of the two sides:
−2A + B + 6C = 0
−2B + 2C = 0
−2C = −12
⇒
A = 21
B = 6
C = 6
⇒ yp(t) = 21 + 6t + 6t2
◮ The General Solutions of the Nonhomogeneous Equation:
y(t) = yp(t) + yc(t) = 21 + 6t + 6t2 + C1e−t + C2e
2
3t.
Nonhomogeneous Linear Equations:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t),
Towards the Rules of Setting Up the Trial Function:
f(t) yp(t)
pN (t) (a polynomial of deg N) A0 + A1t + · · · + AN tN
kert Aert
pN (t)ert (A0 + A1t + · · · + AN tN )ert
· · · · · ·
(to be continued)
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = A cos(2t).
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = A cos(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t)]′′ + [A cos(2t)]′ − 2A cos(2t)
= 3(−4A cos(2t)) − 2A sin(2t) − 2A cos(2t)
= −14A cos(2t) − 2A sin(2t)
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = A cos(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t)]′′ + [A cos(2t)]′ − 2A cos(2t)
= 3(−4A cos(2t)) − 2A sin(2t) − 2A cos(2t)
= −14A cos(2t) − 2A sin(2t)
◮ Compare the coefficients of the two sides:{
−14A = 5−2A = 0
⇒ Impossible!
Example 4: Find a particular solution of 3y′′ + y′ − 2y = 5 cos(2t).
◮ Complementary solutions yc(t):
3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e
2
3t
◮ To find yp(t), set the trial function
yp(t) = A cos(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t)]′′ + [A cos(2t)]′ − 2A cos(2t)
= 3(−4A cos(2t)) − 2A sin(2t) − 2A cos(2t)
= −14A cos(2t) − 2A sin(2t)
◮ Compare the coefficients of the two sides:{
−14A = 5−2A = 0
⇒ Impossible!
The choice of the trial function yp(t) = A cos(2t) wasWRONG!
Example 4 (continued): Find a particular solution of3y′′ + y′ − 2y = 5 cos(2t).
◮ The correct trial function:yp(t) = A cos(2t) + B sin(2t).
Example 4 (continued): Find a particular solution of3y′′ + y′ − 2y = 5 cos(2t).
◮ The correct trial function:yp(t) = A cos(2t) + B sin(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t) + B sin(2t)]′′ + [A cos(2t) + B sin(2t)]′
−2[A cos(2t) + B sin(2t)]
= 3[−4A cos(2t) − 4B sin(2t)] + [−2A sin(2t) + 2B cos(2t)]−2[A cos(2t) + B sin(2t)]
= (−14A + 2B) cos(2t) + (−2A − 14B) sin(2t).
Example 4 (continued): Find a particular solution of3y′′ + y′ − 2y = 5 cos(2t).
◮ The correct trial function:yp(t) = A cos(2t) + B sin(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t) + B sin(2t)]′′ + [A cos(2t) + B sin(2t)]′
−2[A cos(2t) + B sin(2t)]
= 3[−4A cos(2t) − 4B sin(2t)] + [−2A sin(2t) + 2B cos(2t)]−2[A cos(2t) + B sin(2t)]
= (−14A + 2B) cos(2t) + (−2A − 14B) sin(2t).
◮ Compare the coefficients of the two sides:{
−14A + 2B = 5
−2A − 14B = 0⇒
{
A = − 7
20
B = 1
20
Example 4 (continued): Find a particular solution of3y′′ + y′ − 2y = 5 cos(2t).
◮ The correct trial function:yp(t) = A cos(2t) + B sin(2t).
◮ Substitute yp(t) in the nonhomog eq:
5 cos(2t) = 3[A cos(2t) + B sin(2t)]′′ + [A cos(2t) + B sin(2t)]′
−2[A cos(2t) + B sin(2t)]
= 3[−4A cos(2t) − 4B sin(2t)] + [−2A sin(2t) + 2B cos(2t)]−2[A cos(2t) + B sin(2t)]
= (−14A + 2B) cos(2t) + (−2A − 14B) sin(2t).
◮ Compare the coefficients of the two sides:{
−14A + 2B = 5
−2A − 14B = 0⇒
{
A = − 7
20
B = 1
20
◮ A Particular Solution of the Nonhomogeneous Equation:
yp(t) = − 7
20cos(2t) + 1
20sin(2t).
Nonhomogeneous Linear Equations:
a2y′′(t) + a1y
′(t) + a0y(t) = f(t)
Towards the Rules of Setting Up the Trial Function:
f(t) yp(t)
pN (t) (a polynomial of deg N) A0 + A1t + · · · + AN tN
pN(t)ert (A0 + A1t + · · · + AN tN )ert
{
pN (t) cos(ωt)and/or pN(t) sin(ωt)
(A0 + A1t + · · · + AN tN ) cos(ωt)
+(B0 + B1t + · · · + BN tN ) sin(ωt)
{
pN (t)ert cos(ωt)and/or pN (t)ert sin(ωt)
(A0 + A1t + · · · + AN tN )ert cos(ωt)
+(B0 + B1t + · · · + BN tN )ert sin(ωt)
(to be continued)
Example 5: Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2+5 cos(2t)+17e−t cos t+34e−t sin t
Example 5: Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2+5 cos(2t)+17e−t cos t+34e−t sin t
We give two methods.
Example 5: Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2+5 cos(2t)+17e−t cos t+34e−t sin t
We give two methods.
Method 1:Solve 3y′′
1 + y′
1 − 2y1 = 10e4t to get a particular solution y1(t).
Solve 3y′′
2 + y′
2 − 2y2 = −8te−2t to get a particular solution y2(t).
Solve 3y′′
3 + y′
3 − 2y3 = −12t2 to get a particular solution y3(t).
Solve 3y′′
4 + y′
4 − 2y4 = 5 cos(2t) to get a particular solution y4(t).
Solve 3y′′
5 + y′
5 − 2y5 = 17e−t cos t + 34e−t sin t to get a particularsolution y5(t). (Set y5(t) = Ae−t cos t + Be−t sin t)
Example 5: Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2+5 cos(2t)+17e−t cos t+34e−t sin t
We give two methods.
Method 1:Solve 3y′′
1 + y′
1 − 2y1 = 10e4t to get a particular solution y1(t).
Solve 3y′′
2 + y′
2 − 2y2 = −8te−2t to get a particular solution y2(t).
Solve 3y′′
3 + y′
3 − 2y3 = −12t2 to get a particular solution y3(t).
Solve 3y′′
4 + y′
4 − 2y4 = 5 cos(2t) to get a particular solution y4(t).
Solve 3y′′
5 + y′
5 − 2y5 = 17e−t cos t + 34e−t sin t to get a particularsolution y5(t). (Set y5(t) = Ae−t cos t + Be−t sin t)
A particular solution to the original equation:
yp(t) = y1(t) + y2(t) + y3(t) + y4(t) + y5(t)
= 1
5e2t +
(
− 11
8− t
)
e−2t + 21 + 6t + 6t2
− 7
20cos(2t) + 1
20sin(2t) + 7
2e−t cos t − 11
2e−t sin t.
Example 5 (continued): Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2 +5 cos(2t)+17e−t cos t + 34e−t sin t
Method 2:
◮ Set a BIIIIIG trial function:
yp(t) = A0e4t + (A1 + A2t)e
−2t + (A3 + A4t + A5t2)
+[A6 cos(2t) + A7 sin(2t)] + (A8e−t cos t + A9e
−t sin t),
with undetermined coefficients A0, A1, · · · , A9.
Example 5 (continued): Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2 +5 cos(2t)+17e−t cos t + 34e−t sin t
Method 2:
◮ Set a BIIIIIG trial function:
yp(t) = A0e4t + (A1 + A2t)e
−2t + (A3 + A4t + A5t2)
+[A6 cos(2t) + A7 sin(2t)] + (A8e−t cos t + A9e
−t sin t),
with undetermined coefficients A0, A1, · · · , A9.
◮ Substitute this in the original equation.
Example 5 (continued): Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2 +5 cos(2t)+17e−t cos t + 34e−t sin t
Method 2:
◮ Set a BIIIIIG trial function:
yp(t) = A0e4t + (A1 + A2t)e
−2t + (A3 + A4t + A5t2)
+[A6 cos(2t) + A7 sin(2t)] + (A8e−t cos t + A9e
−t sin t),
with undetermined coefficients A0, A1, · · · , A9.
◮ Substitute this in the original equation.
◮ Compare the coefficients of the two sides⇒ Linear equations for A0, A1, · · ·A9.
Example 5 (continued): Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2 +5 cos(2t)+17e−t cos t + 34e−t sin t
Method 2:
◮ Set a BIIIIIG trial function:
yp(t) = A0e4t + (A1 + A2t)e
−2t + (A3 + A4t + A5t2)
+[A6 cos(2t) + A7 sin(2t)] + (A8e−t cos t + A9e
−t sin t),
with undetermined coefficients A0, A1, · · · , A9.
◮ Substitute this in the original equation.
◮ Compare the coefficients of the two sides⇒ Linear equations for A0, A1, · · ·A9.
◮ Solve A0, A1, · · ·A9.
Example 5 (continued): Find a particular solution of
3y′′+y′−2y = 10e4t−8te−2t−12t2 +5 cos(2t)+17e−t cos t + 34e−t sin t
Method 2:
◮ Set a BIIIIIG trial function:
yp(t) = A0e4t + (A1 + A2t)e
−2t + (A3 + A4t + A5t2)
+[A6 cos(2t) + A7 sin(2t)] + (A8e−t cos t + A9e
−t sin t),
with undetermined coefficients A0, A1, · · · , A9.
◮ Substitute this in the original equation.
◮ Compare the coefficients of the two sides⇒ Linear equations for A0, A1, · · ·A9.
◮ Solve A0, A1, · · ·A9.
◮ Finally obtain the particular solution
yp(t) = 1
5e2t +
(
− 11
8− t
)
e−2t + 21 + 6t + 6t2
− 7
20cos(2t) + 1
20sin(2t) + 7
2e−t cos t − 11
2e−t sin t.
(Computational details skipped here.)
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e
2t + C2e−t
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e
2t + C2e−t
◮ To find yp(t), set the trial function
yp(t) = Ae3t + Be2t.
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e
2t + C2e−t
◮ To find yp(t), set the trial function
yp(t) = Ae3t + Be2t.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Be2t)′′ − (Ae3t + Be2t)′ − 2(Ae3t + Be2t)
= (9Ae3t + 4Be2t) − (3Ae3t + 2Be2t) − 2(Ae3t + Be2t)
= 4Ae3t
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e
2t + C2e−t
◮ To find yp(t), set the trial function
yp(t) = Ae3t + Be2t.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Be2t)′′ − (Ae3t + Be2t)′ − 2(Ae3t + Be2t)
= (9Ae3t + 4Be2t) − (3Ae3t + 2Be2t) − 2(Ae3t + Be2t)
= 4Ae3t
◮ Compare the coefficients of the two sides:{
4A = 360 = 2
⇒ Impossible!
Example 6: Find general solutions of y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e
2t + C2e−t
◮ To find yp(t), set the trial function
yp(t) = Ae3t + Be2t.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Be2t)′′ − (Ae3t + Be2t)′ − 2(Ae3t + Be2t)
= (9Ae3t + 4Be2t) − (3Ae3t + 2Be2t) − 2(Ae3t + Be2t)
= 4Ae3t
◮ Compare the coefficients of the two sides:{
4A = 360 = 2
⇒ Impossible!
The trial function yp(t) = Ae3t + Be2t was BAD!
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
◮ The Be2t part of the trial function satisfies the homog eq.
That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
◮ The Be2t part of the trial function satisfies the homog eq.
That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.
◮ In other words, when plugged in the nonhomog equation,this Be2t produces many terms,but the sum of those terms will simplify to zero!
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
◮ The Be2t part of the trial function satisfies the homog eq.
That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.
◮ In other words, when plugged in the nonhomog equation,this Be2t produces many terms,but the sum of those terms will simplify to zero!
◮ Thus, impossible to balance the two sides of the nonhomogequation.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
◮ The Be2t part of the trial function satisfies the homog eq.
That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.
◮ In other words, when plugged in the nonhomog equation,this Be2t produces many terms,but the sum of those terms will simplify to zero!
◮ Thus, impossible to balance the two sides of the nonhomogequation.
◮ In short, the failure was due to the fact that
yp(t) has overlap(s) with yc(t).
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc = C1e2t + C2e
−t
◮ The BAD trial function: yp(t) = Ae3t + Be2t.
◮ The Reason of the Failure:
◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.
◮ The Be2t part of the trial function satisfies the homog eq.
That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.
◮ In other words, when plugged in the nonhomog equation,this Be2t produces many terms,but the sum of those terms will simplify to zero!
◮ Thus, impossible to balance the two sides of the nonhomogequation.
◮ In short, the failure was due to the fact that
yp(t) has overlap(s) with yc(t).
◮ This kind of cases are called resonance.
The term 2e2t in f(t) is called a resonant term.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.◮ The Method of Correction: In the naive trial function,
multiply the bad term Be2t by tk , where k is the smallest positive integer to
ensure that yp does not overlap yc.
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.◮ The Method of Correction: In the naive trial function,
multiply the bad term Be2t by tk , where k is the smallest positive integer to
ensure that yp does not overlap yc.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Bte2t)′′ − (Ae3t + Bte2t)′ − 2(Ae3t + Bte2t)
= (9Ae3t + 4Be2t + 4Bte2t) − (3Ae3t + Be2t + 2Bte2t)
−2(Ae3t + Bte2t)
= 4Ae3t + 3Be2t
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.◮ The Method of Correction: In the naive trial function,
multiply the bad term Be2t by tk , where k is the smallest positive integer to
ensure that yp does not overlap yc.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Bte2t)′′ − (Ae3t + Bte2t)′ − 2(Ae3t + Bte2t)
= (9Ae3t + 4Be2t + 4Bte2t) − (3Ae3t + Be2t + 2Bte2t)
−2(Ae3t + Bte2t)
= 4Ae3t + 3Be2t
◮ Compare the coefficients of the two sides:{
4A = 363B = 2
⇒
{
A = 9B = 2/3
⇒ yp(t) = 9e3t + 2
3te2t
Example 6 (continued): y′′ − y′ − 2y = 36e3t + 2e2t.
◮ Complementary solutions: yc(t) = C1e2t + C2e
−t
◮ The NAIVE trial function: yp(t) = Ae3t + Be2t.This failed, since it has a term Be2t overlapping yc(t).
◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.◮ The Method of Correction: In the naive trial function,
multiply the bad term Be2t by tk , where k is the smallest positive integer to
ensure that yp does not overlap yc.
◮ Substitute yp(t) in the nonhomog eq:
36e3t + 2e2t = (Ae3t + Bte2t)′′ − (Ae3t + Bte2t)′ − 2(Ae3t + Bte2t)
= (9Ae3t + 4Be2t + 4Bte2t) − (3Ae3t + Be2t + 2Bte2t)
−2(Ae3t + Bte2t)
= 4Ae3t + 3Be2t
◮ Compare the coefficients of the two sides:{
4A = 363B = 2
⇒
{
A = 9B = 2/3
⇒ yp(t) = 9e3t + 2
3te2t
◮ The general solutionsy(t) = yp(t) + yc(t) = 9e3t + 2
3te2t + C1e
2t + C2e−t
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
◮ The modified trial function:yp(t) = ae4t + t (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
◮ The modified trial function:yp(t) = ae4t + t2(b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
◮ The modified trial function:yp(t) = ae4t + t2(b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
◮ Substitute this correct yp(t) in the nonhomog eq:
9e4t + (3 − 2t − t2)e−2t + cos t = y′′
p + 4y′
p + 4yp = · · · · · ·
= 36ae4t + (2b0 + 6b1t + 12b2)e−2t + (3A + 4B) cos t + (−4A + 3B) sin t
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
◮ The modified trial function:yp(t) = ae4t + t2(b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
◮ Substitute this correct yp(t) in the nonhomog eq:
9e4t + (3 − 2t − t2)e−2t + cos t = y′′
p + 4y′
p + 4yp = · · · · · ·
= 36ae4t + (2b0 + 6b1t + 12b2)e−2t + (3A + 4B) cos t + (−4A + 3B) sin t
◮ Compare the coefficients of the two sides:
36a = 92b0 = 3, 6b1 = −2, 12b2 = −13A + 4B = 1, −4A + 3B = 0
⇒
a = 1
4
b0 = 3
2, b1 = − 1
3, b2 = − 1
12
A = 3
25, B = 4
25
Example 7: Find a particular solution ofy′′ + 4y′ + 4y = 9e4t + (3 − 2t − t2)e−2t + cos t.
◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e
−2t + C2te−2t
◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e
−2t overlaps yc(t).
◮ The modified trial function:yp(t) = ae4t + t2(b0 + b1t + b2t
2)e−2t + (A cos t + B sin t).
◮ Substitute this correct yp(t) in the nonhomog eq:
9e4t + (3 − 2t − t2)e−2t + cos t = y′′
p + 4y′
p + 4yp = · · · · · ·
= 36ae4t + (2b0 + 6b1t + 12b2)e−2t + (3A + 4B) cos t + (−4A + 3B) sin t
◮ Compare the coefficients of the two sides:
36a = 92b0 = 3, 6b1 = −2, 12b2 = −13A + 4B = 1, −4A + 3B = 0
⇒
a = 1
4
b0 = 3
2, b1 = − 1
3, b2 = − 1
12
A = 3
25, B = 4
25
◮ The particular solutionyp(t) = 1
4e4t + t2
(
3
2− 1
3t − 1
12t2
)
e−2t + 3
25cos t + 4
25sin t.
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
◮ Complementary solutions: λ2 + 2λ + 10 = 0 ⇒ λ1,2 = −1 ± 3i⇒ yc = C1e
−t cos(3t) + C2e−t sin(3t)
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
◮ Complementary solutions: λ2 + 2λ + 10 = 0 ⇒ λ1,2 = −1 ± 3i⇒ yc = C1e
−t cos(3t) + C2e−t sin(3t)
◮The trial function: yp(t) = ae−t + [b1 cos(3t) + b2 sin(3t)]
+[Ate−t cos(3t) + Bte−t sin(3t)]
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
◮ Complementary solutions: λ2 + 2λ + 10 = 0 ⇒ λ1,2 = −1 ± 3i⇒ yc = C1e
−t cos(3t) + C2e−t sin(3t)
◮The trial function: yp(t) = ae−t + [b1 cos(3t) + b2 sin(3t)]
+[Ate−t cos(3t) + Bte−t sin(3t)]
◮ Substitute yp(t) in the nonhomog eq and simplify:
e−t + cos(3t) + e−t sin(3t) = y′′
p + 2y′
p + 10yp = · · · · · ·
= 9ae−t + (b1 + 6b2) cos(3t) + (−6b1 + b2) sin(3t)
+6Be−t cos(3t) − 6Ae−t sin(3t).
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
◮ Complementary solutions: λ2 + 2λ + 10 = 0 ⇒ λ1,2 = −1 ± 3i⇒ yc = C1e
−t cos(3t) + C2e−t sin(3t)
◮The trial function: yp(t) = ae−t + [b1 cos(3t) + b2 sin(3t)]
+[Ate−t cos(3t) + Bte−t sin(3t)]
◮ Substitute yp(t) in the nonhomog eq and simplify:
e−t + cos(3t) + e−t sin(3t) = y′′
p + 2y′
p + 10yp = · · · · · ·
= 9ae−t + (b1 + 6b2) cos(3t) + (−6b1 + b2) sin(3t)
+6Be−t cos(3t) − 6Ae−t sin(3t).
◮ Compare the coefficients of the two sides:
9a = 1b1 + 6b2 = 1, −6b1 + b2 = 06B = 0, −6A = 1
⇒
a = 1/9b1 = 1/37, b2 = 6/37A = −1/6, B = 0
Example 8: Find a particular solution ofy′′ + 2y′ + 10y = e−t + cos(3t) + e−t sin(3t).
◮ Complementary solutions: λ2 + 2λ + 10 = 0 ⇒ λ1,2 = −1 ± 3i⇒ yc = C1e
−t cos(3t) + C2e−t sin(3t)
◮The trial function: yp(t) = ae−t + [b1 cos(3t) + b2 sin(3t)]
+[Ate−t cos(3t) + Bte−t sin(3t)]
◮ Substitute yp(t) in the nonhomog eq and simplify:
e−t + cos(3t) + e−t sin(3t) = y′′
p + 2y′
p + 10yp = · · · · · ·
= 9ae−t + (b1 + 6b2) cos(3t) + (−6b1 + b2) sin(3t)
+6Be−t cos(3t) − 6Ae−t sin(3t).
◮ Compare the coefficients of the two sides:
9a = 1b1 + 6b2 = 1, −6b1 + b2 = 06B = 0, −6A = 1
⇒
a = 1/9b1 = 1/37, b2 = 6/37A = −1/6, B = 0
◮ The particular solutionyp(t) = 1
10e−2t + 1
37cos(3t) + 6
37sin(3t) − 1
6te−t cos(3t).
Summary of the Method of Undetermined Coefficients
Nonhomog Linear Equations: a2y′′(t) + a1y
′(t) + a0y(t) = f(t)
How to set up the trial function?
f(t) yp(t)
pN (t) (a polynomial of deg N) A0 + A1t + · · · + AN tN
pN(t)ert (A0 + A1t + · · · + AN tN )ert
{
pN (t) cos(ωt)and/or pN(t) sin(ωt)
(A0 + A1t + · · · + AN tN ) cos(ωt)
+(B0 + B1t + · · · + BN tN ) sin(ωt)
{
pN (t)ert cos(ωt)and/or pN (t)ert sin(ωt)
(A0 + A1t + · · · + AN tN )ert cos(ωt)
+(B0 + B1t + · · · + BN tN )ert sin(ωt)
In the case of resonance:• First pick a naive trial function as in the above table.• Then multiply the resonant term(s) by tk, where k is the smallest
positive integer to ensure that yp does not overlap yc.
Bad News
Bad News
◮ The method of undetermined coefficients does NOOOOOOTwork, when the equation has variable coefficients :
a2(t)y′′ + a1(t)y
′ + a0(t)y = f(t)
Example: (t−1)y′′− ty′ + y = e2t cannot be solved by the m.u.c.
Bad News
◮ The method of undetermined coefficients does NOOOOOOTwork, when the equation has variable coefficients :
a2(t)y′′ + a1(t)y
′ + a0(t)y = f(t)
Example: (t−1)y′′− ty′ + y = e2t cannot be solved by the m.u.c.
◮ Even for the equations of constant coefficients:
a2y′′ + a1y
′ + a0y = f(t),
the m.u.c. does NOOOOOOT always work.
It only works when f(t) is a linear combination of the functionsthat appear in the table of the last page.
Examples for which the m.u.c. fails:
y′′ + y = tan t, y′′ + 2y′ + y = et2 , y′′ − y = 1
1+t, · · ·
Bad News
◮ The method of undetermined coefficients does NOOOOOOTwork, when the equation has variable coefficients :
a2(t)y′′ + a1(t)y
′ + a0(t)y = f(t)
Example: (t−1)y′′− ty′ + y = e2t cannot be solved by the m.u.c.
◮ Even for the equations of constant coefficients:
a2y′′ + a1y
′ + a0y = f(t),
the m.u.c. does NOOOOOOT always work.
It only works when f(t) is a linear combination of the functionsthat appear in the table of the last page.
Examples for which the m.u.c. fails:
y′′ + y = tan t, y′′ + 2y′ + y = et2 , y′′ − y = 1
1+t, · · ·
Good News
◮ There is a more general method, the variation of parameters,that can solve any nonhomog linear differential equation,as long as yc has been provided/prepared.