Second Order Nonhomogeneous Linear Differential …people.math.gatech.edu/~xchen/teach/ode/2nd_ord_nonhomog...Second Order Nonhomogeneous Linear ... are solutions of the homogeneous

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).


a2y′′(t) + a1y

′(t) + a0y(t) = f(t),





a2y′′

c (t) + a1y′

c(t) + a0yc(t) = 0.




a2y′′(t) + a1y

′(t) + a0y(t) = f(t),





a2y′′

c (t) + a1y′

c(t) + a0yc(t) = 0.



◮ What is this note about? The Method ofUndetermined Coefficients:


a2y′′(t) + a1y

′(t) + a0y(t) = f(t),





a2y′′

c (t) + a1y′

c(t) + a0yc(t) = 0.



◮ What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.


a2y′′(t) + a1y

′(t) + a0y(t) = f(t),





a2y′′

c (t) + a1y′

c(t) + a0yc(t) = 0.



◮ 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.


◮ 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




3y′′

c + y′


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.




3y′′

c + y′


3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t




◮ Substitute yp(t) in the nonhomog eq:

3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) − 2ae4t

= 50ae4t




3y′′

c + y′


3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t





3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) − 2ae4t

= 50ae4t

◮ Compare the coefficients of the two sides:

50a = 10 ⇒ a =1

5




3y′′

c + y′


3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t





3(ae4t)′′ + (ae4t)′ − 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) − 2ae4t

= 50ae4t


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.




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




5e4t + C1e

−t + C2e2

3t.


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




5e4t + C1e

−t + C2e2

3t.


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.


◮ Complementary solutions yc(t):

3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = Ate−2t.



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = Ate−2t.


−8te−2t = 3(Ate−2t)′′ + (Ate−2t)′ − 2Ate−2t

= 3(−4Ae−2t + 4Ate−2t) + (Ae−2t − 2Ate−2t) − 2Ate−2t

= −11Ae−2t + 8Ate−2t



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = Ate−2t.




= −11Ae−2t + 8Ate−2t

◮ Compare the coefficients of the two sides:{

−11A = 08A = −8

⇒ Impossible!



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = Ate−2t.




= −11Ae−2t + 8Ate−2t


−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.




◮ The correct trial function:yp(t) = (A + Bt)e−2t.






−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







= 3(4A − 4B + 4Bt)e−2t + (−2A + B − 2Bt)e−2t

+(−2A − 2Bt)e−2t

= (8A − 11B)e−2t + 8Bte−2t


8A − 11B = 0

8B = −8⇒

{

A = − 11

8

B = −1⇒ yp(t) =

(

− 11

8− t

)

e−2t







= 3(4A − 4B + 4Bt)e−2t + (−2A + B − 2Bt)e−2t

+(−2A − 2Bt)e−2t

= (8A − 11B)e−2t + 8Bte−2t


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.



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t



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.



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!



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


◮ The correct trial function:yp(t) = A + Bt + Ct2.



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t




−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)

= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t




−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)

= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2


−2A + B + 6C = 0

−2B + 2C = 0

−2C = −12

⇒

A = 21

B = 6

C = 6

⇒ yp(t) = 21 + 6t + 6t2



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t




−12t2 = 3(2C) + (B + 2Ct) − 2(A + Bt + Ct2)

= (−2A + B + 6C) + (−2B + 2C)t − 2Ct2


−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.


a2y′′(t) + a1y

′(t) + a0y(t) = f(t),


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).



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = A cos(2t).



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = A cos(2t).


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)



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = A cos(2t).




= −14A cos(2t) − 2A sin(2t)


−14A = 5−2A = 0

⇒ Impossible!



3λ2 + λ − 2 = 0 ⇒ λ1 = −1, λ2 = 2/3 ⇒ yc = C1e−t + C2e

2

3t


yp(t) = A cos(2t).




= −14A cos(2t) − 2A sin(2t)


−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).




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).







= (−14A + 2B) cos(2t) + (−2A − 14B) sin(2t).


−14A + 2B = 5

−2A − 14B = 0⇒

{

A = − 7

20

B = 1

20







= (−14A + 2B) cos(2t) + (−2A − 14B) sin(2t).


−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).


a2y′′(t) + a1y

′(t) + a0y(t) = f(t)


f(t) yp(t)


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



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)




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.



Method 2:


yp(t) = A0e4t + (A1 + A2t)e

−2t + (A3 + A4t + A5t2)


−t sin t),


◮ Substitute this in the original equation.



Method 2:


yp(t) = A0e4t + (A1 + A2t)e

−2t + (A3 + A4t + A5t2)


−t sin t),



◮ Compare the coefficients of the two sides⇒ Linear equations for A0, A1, · · ·A9.



Method 2:


yp(t) = A0e4t + (A1 + A2t)e

−2t + (A3 + A4t + A5t2)


−t sin t),




◮ Solve A0, A1, · · ·A9.



Method 2:


yp(t) = A0e4t + (A1 + A2t)e

−2t + (A3 + A4t + A5t2)


−t sin t),




◮ 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.


◮ Complementary solutions yc(t):λ2 − λ − 2 = 0 ⇒ λ1 = 2, λ2 = −1 ⇒ yc = C1e

2t + C2e−t



2t + C2e−t


yp(t) = Ae3t + Be2t.



2t + C2e−t




36e3t + 2e2t = (Ae3t + Be2t)′′ − (Ae3t + Be2t)′ − 2(Ae3t + Be2t)

= (9Ae3t + 4Be2t) − (3Ae3t + 2Be2t) − 2(Ae3t + Be2t)

= 4Ae3t



2t + C2e−t






= 4Ae3t


4A = 360 = 2

⇒ Impossible!



2t + C2e−t






= 4Ae3t


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.



−t


◮ The Reason of the Failure:



−t



◮ When yp(t) is plugged in the nonhomog eq, we wish the lefthand side would match the right hand side 36e3t + 2e2t.



−t




◮ The Be2t part of the trial function satisfies the homog eq.

That is, (Be2t)′′ − (Be2t)′ − 2Be2t = 0.



−t






◮ In other words, when plugged in the nonhomog equation,this Be2t produces many terms,but the sum of those terms will simplify to zero!



−t







◮ Thus, impossible to balance the two sides of the nonhomogequation.



−t








◮ In short, the failure was due to the fact that

yp(t) has overlap(s) with yc(t).



−t








◮ 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.


◮ 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).



−t


◮ The CORRECT trial function: yp(t) = Ae3t + Bte2t.



−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.



−t






36e3t + 2e2t = (Ae3t + Bte2t)′′ − (Ae3t + Bte2t)′ − 2(Ae3t + Bte2t)

= (9Ae3t + 4Be2t + 4Bte2t) − (3Ae3t + Be2t + 2Bte2t)

−2(Ae3t + Bte2t)

= 4Ae3t + 3Be2t



−t








−2(Ae3t + Bte2t)

= 4Ae3t + 3Be2t


4A = 363B = 2

⇒

{

A = 9B = 2/3

⇒ yp(t) = 9e3t + 2

3te2t



−t








−2(Ae3t + Bte2t)

= 4Ae3t + 3Be2t


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.


◮ Complementary solutions:λ2 + 4λ + 4 = 0 ⇒ λ1 = λ2 = −2 ⇒ yc = C1e

−2t + C2te−2t



−2t + C2te−2t

◮ The trial function:yp(t) = ae4t + (b0 + b1t + b2t

2)e−2t + (A cos t + B sin t).



−2t + C2te−2t


2)e−2t + (A cos t + B sin t).This would fail, since the part (b0 + b1t)e

−2t overlaps yc(t).



−2t + C2te−2t




◮ The modified trial function:yp(t) = ae4t + t (b0 + b1t + b2t




−2t + C2te−2t




◮ The modified trial function:yp(t) = ae4t + t2(b0 + b1t + b2t




−2t + C2te−2t






◮ 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



−2t + C2te−2t







9e4t + (3 − 2t − t2)e−2t + cos t = y′′

p + 4y′

p + 4yp = · · · · · ·



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



−2t + C2te−2t







9e4t + (3 − 2t − t2)e−2t + cos t = y′′

p + 4y′

p + 4yp = · · · · · ·



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).


◮ 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).







e−t + cos(3t) + e−t sin(3t) = y′′

p + 2y′

p + 10yp = · · · · · ·




9a = 1b1 + 6b2 = 1, −6b1 + b2 = 06B = 0, −6A = 1

⇒

a = 1/9b1 = 1/37, b2 = 6/37A = −1/6, B = 0







e−t + cos(3t) + e−t sin(3t) = y′′

p + 2y′

p + 10yp = · · · · · ·




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)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


a2(t)y′′ + a1(t)y

′ + a0(t)y = f(t)


◮ 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


a2(t)y′′ + a1(t)y

′ + a0(t)y = f(t)


◮ 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.

Documents

Second Order Nonhomogeneous Linear Differential …people.math.gatech.edu/~xchen/teach/ode/2nd_ord_nonhomog...Second Order Nonhomogeneous Linear ... are solutions of the homogeneous