Summary of Solving Linear, Constant-Coefficient Recurrence ...
Transcript of Summary of Solving Linear, Constant-Coefficient Recurrence ...
Summary of Solving Linear,Constant-Coefficient Recurrence Relations
Ioan Despi
University of New England
September 27, 2013
Outline
1 The Technique
2 Homogeneous Case
3 Non-Homogeneous Case
4 Examples
Ioan Despi β AMTH140 2 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:
I any answer which works is the correct answer since there is only onecorrect answer (this can be proved!).
There are two parts of the total solution:
1 The homogeneous part of the solution depends only on what is on the leftof the recurrence relation.
2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:
I any answer which works is the correct answer since there is only onecorrect answer (this can be proved!).
There are two parts of the total solution:
1 The homogeneous part of the solution depends only on what is on the leftof the recurrence relation.
2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one
correct answer (this can be proved!).
There are two parts of the total solution:
1 The homogeneous part of the solution depends only on what is on the leftof the recurrence relation.
2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one
correct answer (this can be proved!).
There are two parts of the total solution:
1 The homogeneous part of the solution depends only on what is on the leftof the recurrence relation.
2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one
correct answer (this can be proved!).
There are two parts of the total solution:1 The homogeneous part of the solution depends only on what is on the left
of the recurrence relation.
2 The particular part of the total solution depends on what is in RHS andhas the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one
correct answer (this can be proved!).
There are two parts of the total solution:1 The homogeneous part of the solution depends only on what is on the left
of the recurrence relation.2 The particular part of the total solution depends on what is in RHS and
has the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
The technique is based on method of characteristic equations.
It relies on the principle of uniqueness:I any answer which works is the correct answer since there is only one
correct answer (this can be proved!).
There are two parts of the total solution:1 The homogeneous part of the solution depends only on what is on the left
of the recurrence relation.2 The particular part of the total solution depends on what is in RHS and
has the same form as RHS.
We calculate the two parts separately and add them to form the totalsolution.
Ioan Despi β AMTH140 3 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.
I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.
I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).
I The solution will contain one or more undetermined coefficients whosevalues cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.
I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.
I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.
I This step does not produce any additional undetermined coefficients, nordoes it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.I This step does not produce any additional undetermined coefficients, nor
does it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.I This step does not produce any additional undetermined coefficients, nor
does it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
The Technique
There are four steps in the process:
(S1.) Find the homogeneous solution to the homogeneous equation (the onethat results when you set RHS=0).
I If RHS is already zero, skip next two steps and go to (S4).I The solution will contain one or more undetermined coefficients whose
values cannot be determined until (S4).
(S2.) Find the particular solution by guessing a form similar to RHS.I This step does not produce any additional undetermined coefficients, nor
does it eliminate those from (S1).
(S3.) Combine (add) the homogeneous and particular solutions.
(S4.) Use initial conditions to eliminate the undetermined constants from (S1).
Ioan Despi β AMTH140 4 of 12
Homogeneous Case
1 First solve the characteristic equation.
(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Homogeneous Case
1 First solve the characteristic equation.(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Homogeneous Case
1 First solve the characteristic equation.(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Homogeneous Case
1 First solve the characteristic equation.(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Homogeneous Case
1 First solve the characteristic equation.(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Homogeneous Case
1 First solve the characteristic equation.(a) Distinct roots
Example π = 1, π = 2
Solution ππ = π΄(1)π +π΅(2)π = π΄+π΅2π
(b) Repeated roots
Example π = 3 with multiplicity 2
Solution ππ = (π΄π+π΅)3π
Example π = 2 with multiplicity 3
Solution ππ = (π΄π2 +π΅π+ πΆ)2π
Example π = 2 with multiplicity 1
π = β1 with multiplicity 2
Solution ππ = π΄2π + (π΅π+ πΆ)(β1)π
2 If there are initial conditions, use them to find the values for theconstants π΄,π΅, etc.
Ioan Despi β AMTH140 5 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.
If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.
(a) Solution to the homogeneous case not like the right hand side of therecurrence relation.
F Try substituting something that looks like the right hand side (RHS).Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.
Next find a particular solution, π£π, to the non-homogeneous case.
(a) Solution to the homogeneous case not like the right hand side of therecurrence relation.
F Try substituting something that looks like the right hand side (RHS).Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.
(a) Solution to the homogeneous case not like the right hand side of therecurrence relation.
F Try substituting something that looks like the right hand side (RHS).Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.
F Try substituting something that looks like the right hand side (RHS).Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅
Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)
Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.
(i) Distinct roots: Try a solution with an extra π.Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.(i) Distinct roots: Try a solution with an extra π.
Example π = 2 and RHS = 2π, try π£π = πΆπ2π.
(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 formultiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.(i) Distinct roots: Try a solution with an extra π.
Example π = 2 and RHS = 2π, try π£π = πΆπ2π.(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 for
multiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.(i) Distinct roots: Try a solution with an extra π.
Example π = 2 and RHS = 2π, try π£π = πΆπ2π.(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 for
multiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Solve the characteristic equation and obtain a solution to thehomogeneous case as in 1. Call it π’π.If there are initial conditions leave these for the moment.Next find a particular solution, π£π, to the non-homogeneous case.(a) Solution to the homogeneous case not like the right hand side of the
recurrence relation.F Try substituting something that looks like the right hand side (RHS).
Example RHS = 2π, try π£π = πΆ2π
Example RHS = 4π, try π£π = π΄π+π΅Example RHS = 3ππ, try π£π = 3π(π΄π+π΅)Example RHS = π2, try π£π = π΄π2 +π΅π+ πΆ
(b) Solution to the homogeneous case similar the the RHS of the recurrencerelation.(i) Distinct roots: Try a solution with an extra π.
Example π = 2 and RHS = 2π, try π£π = πΆπ2π.(ii) Repeat Roots: Try a solution with an extra π2 for multiplicity 2, π3 for
multiplicity 3 etc.
Example π = 3 with multiplicity 2 and RHS = 3π, try
π£π = πΆπ23π
Example π = 1 with multiplicity 3 and RHS = 4 = 4Γ 1π, try
π£π = πΆπ3 Β· 13 = πΆπ3
Ioan Despi β AMTH140 6 of 12
Non-Homogeneous Case
Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.
Then find values for the constants and put these values back into theexpressions for π£π.
Now add π’π and π£π to obtain the general solution. That is,
ππ = π’π + π£π.
If there are initial conditions, substitute them in now to find the values ofthe constants that were part of π£π.
Finally, check your answer by substituting it into the original recurrencerelation.
Ioan Despi β AMTH140 7 of 12
Non-Homogeneous Case
Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.
Then find values for the constants and put these values back into theexpressions for π£π.
Now add π’π and π£π to obtain the general solution. That is,
ππ = π’π + π£π.
If there are initial conditions, substitute them in now to find the values ofthe constants that were part of π£π.
Finally, check your answer by substituting it into the original recurrencerelation.
Ioan Despi β AMTH140 7 of 12
Non-Homogeneous Case
Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.
Then find values for the constants and put these values back into theexpressions for π£π.
Now add π’π and π£π to obtain the general solution. That is,
ππ = π’π + π£π.
If there are initial conditions, substitute them in now to find the values ofthe constants that were part of π£π.
Finally, check your answer by substituting it into the original recurrencerelation.
Ioan Despi β AMTH140 7 of 12
Non-Homogeneous Case
Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.
Then find values for the constants and put these values back into theexpressions for π£π.
Now add π’π and π£π to obtain the general solution. That is,
ππ = π’π + π£π.
If there are initial conditions, substitute them in now to find the values ofthe constants that were part of π£π.
Finally, check your answer by substituting it into the original recurrencerelation.
Ioan Despi β AMTH140 7 of 12
Non-Homogeneous Case
Once you have decided what a possible solution to the nonhomogeneouscase could be, substitute it into the original recurrence relation.
Then find values for the constants and put these values back into theexpressions for π£π.
Now add π’π and π£π to obtain the general solution. That is,
ππ = π’π + π£π.
If there are initial conditions, substitute them in now to find the values ofthe constants that were part of π£π.
Finally, check your answer by substituting it into the original recurrencerelation.
Ioan Despi β AMTH140 7 of 12
Examples
Example
Find the general solution of the recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0 .
Solution.
The associated characteristic equation
π2 β 5π + 6 = 0
has two (distinct) roots π1 = 2 and π2 = 3.
Hence the general solution of the recurrence relation is
ππ = π΄2π + π΅3π, π β₯ 0 ,
where π΄ and π΅ are arbitrary constants.
Ioan Despi β AMTH140 8 of 12
Examples
Example
Find the general solution of the recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0 .
Solution.
The associated characteristic equation
π2 β 5π + 6 = 0
has two (distinct) roots π1 = 2 and π2 = 3.
Hence the general solution of the recurrence relation is
ππ = π΄2π + π΅3π, π β₯ 0 ,
where π΄ and π΅ are arbitrary constants.
Ioan Despi β AMTH140 8 of 12
Examples
Example
Find the general solution of the recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0 .
Solution.
The associated characteristic equation
π2 β 5π + 6 = 0
has two (distinct) roots π1 = 2 and π2 = 3.
Hence the general solution of the recurrence relation is
ππ = π΄2π + π΅3π, π β₯ 0 ,
where π΄ and π΅ are arbitrary constants.
Ioan Despi β AMTH140 8 of 12
Examples
Example
Find a particular solution of the previous recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0
such that it satisfies the initial conditions π0 = 2 and π1 = 3.
Solution.
From the general solution obtained in the previous example, the initialconditions give rise to the following two equations
π0 = π΄ + π΅ = 2 , π1 = π΄Γ 2 + π΅ Γ 3 = 3
Solving the above 2 equations we obtain π΄ = 3 and π΅ = β1.Hence the particular solution satisfying the initial conditions is
ππ = 3 Γ 2π β 3π, π β₯ 0
Ioan Despi β AMTH140 9 of 12
Examples
Example
Find a particular solution of the previous recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0
such that it satisfies the initial conditions π0 = 2 and π1 = 3.
Solution.
From the general solution obtained in the previous example, the initialconditions give rise to the following two equations
π0 = π΄ + π΅ = 2 , π1 = π΄Γ 2 + π΅ Γ 3 = 3
Solving the above 2 equations we obtain π΄ = 3 and π΅ = β1.
Hence the particular solution satisfying the initial conditions is
ππ = 3 Γ 2π β 3π, π β₯ 0
Ioan Despi β AMTH140 9 of 12
Examples
Example
Find a particular solution of the previous recurrence relation
ππ+2 β 5ππ+1 + 6ππ = 0, π β₯ 0
such that it satisfies the initial conditions π0 = 2 and π1 = 3.
Solution.
From the general solution obtained in the previous example, the initialconditions give rise to the following two equations
π0 = π΄ + π΅ = 2 , π1 = π΄Γ 2 + π΅ Γ 3 = 3
Solving the above 2 equations we obtain π΄ = 3 and π΅ = β1.Hence the particular solution satisfying the initial conditions is
ππ = 3 Γ 2π β 3π, π β₯ 0
Ioan Despi β AMTH140 9 of 12
Examples
Example
Find the general solution for the recurrence relation
π€π+2 β 4π€π+1 + 4π€π = 2π+2 , π β₯ 0
Solution.
The associated characteristic equation π2 β 4π + 4 = (πβ 2)2 = 0 has adouble root π1 = 2.Hence the general solution of the corresponding homogeneous problem isπ’π = 2π(π΄ + π΅π) for π β₯ 0, where π΄ and π΅ are arbitrary constants.Since the nonhomogeneous term is 2π+2 = 4 Γ 2π, a particular solution π£πwill take the form π£π = πΆπ22π because 2 is a double root of theassociated characteristic equation.Hence, substituting π£π into the nonhomogeneous recurrence relationπ£π+2 β 4π£π+1 + 4π£π = 2π+2, we obtain
πΆ[οΈ4(π + 2)2 β 4(π + 1)2 Γ 2 + 4π2
]οΈ= 4
which then simplifies to just 2πΆ = 1.
Ioan Despi β AMTH140 10 of 12
Examples
Example
Find the general solution for the recurrence relation
π€π+2 β 4π€π+1 + 4π€π = 2π+2 , π β₯ 0
Solution.
The associated characteristic equation π2 β 4π + 4 = (πβ 2)2 = 0 has adouble root π1 = 2.
Hence the general solution of the corresponding homogeneous problem isπ’π = 2π(π΄ + π΅π) for π β₯ 0, where π΄ and π΅ are arbitrary constants.Since the nonhomogeneous term is 2π+2 = 4 Γ 2π, a particular solution π£πwill take the form π£π = πΆπ22π because 2 is a double root of theassociated characteristic equation.Hence, substituting π£π into the nonhomogeneous recurrence relationπ£π+2 β 4π£π+1 + 4π£π = 2π+2, we obtain
πΆ[οΈ4(π + 2)2 β 4(π + 1)2 Γ 2 + 4π2
]οΈ= 4
which then simplifies to just 2πΆ = 1.
Ioan Despi β AMTH140 10 of 12
Examples
Example
Find the general solution for the recurrence relation
π€π+2 β 4π€π+1 + 4π€π = 2π+2 , π β₯ 0
Solution.
The associated characteristic equation π2 β 4π + 4 = (πβ 2)2 = 0 has adouble root π1 = 2.Hence the general solution of the corresponding homogeneous problem isπ’π = 2π(π΄ + π΅π) for π β₯ 0, where π΄ and π΅ are arbitrary constants.
Since the nonhomogeneous term is 2π+2 = 4 Γ 2π, a particular solution π£πwill take the form π£π = πΆπ22π because 2 is a double root of theassociated characteristic equation.Hence, substituting π£π into the nonhomogeneous recurrence relationπ£π+2 β 4π£π+1 + 4π£π = 2π+2, we obtain
πΆ[οΈ4(π + 2)2 β 4(π + 1)2 Γ 2 + 4π2
]οΈ= 4
which then simplifies to just 2πΆ = 1.
Ioan Despi β AMTH140 10 of 12
Examples
Example
Find the general solution for the recurrence relation
π€π+2 β 4π€π+1 + 4π€π = 2π+2 , π β₯ 0
Solution.
The associated characteristic equation π2 β 4π + 4 = (πβ 2)2 = 0 has adouble root π1 = 2.Hence the general solution of the corresponding homogeneous problem isπ’π = 2π(π΄ + π΅π) for π β₯ 0, where π΄ and π΅ are arbitrary constants.Since the nonhomogeneous term is 2π+2 = 4 Γ 2π, a particular solution π£πwill take the form π£π = πΆπ22π because 2 is a double root of theassociated characteristic equation.
Hence, substituting π£π into the nonhomogeneous recurrence relationπ£π+2 β 4π£π+1 + 4π£π = 2π+2, we obtain
πΆ[οΈ4(π + 2)2 β 4(π + 1)2 Γ 2 + 4π2
]οΈ= 4
which then simplifies to just 2πΆ = 1.
Ioan Despi β AMTH140 10 of 12
Examples
Example
Find the general solution for the recurrence relation
π€π+2 β 4π€π+1 + 4π€π = 2π+2 , π β₯ 0
Solution.
The associated characteristic equation π2 β 4π + 4 = (πβ 2)2 = 0 has adouble root π1 = 2.Hence the general solution of the corresponding homogeneous problem isπ’π = 2π(π΄ + π΅π) for π β₯ 0, where π΄ and π΅ are arbitrary constants.Since the nonhomogeneous term is 2π+2 = 4 Γ 2π, a particular solution π£πwill take the form π£π = πΆπ22π because 2 is a double root of theassociated characteristic equation.Hence, substituting π£π into the nonhomogeneous recurrence relationπ£π+2 β 4π£π+1 + 4π£π = 2π+2, we obtain
πΆ[οΈ4(π + 2)2 β 4(π + 1)2 Γ 2 + 4π2
]οΈ= 4
which then simplifies to just 2πΆ = 1.
Ioan Despi β AMTH140 10 of 12
Examples
Hence we have πΆ = 12 and π£π = π22πβ1.
The general solution π€π of the nonhomogeneous recurrence relation isthus π€π = π’π + π£π, and hence takes the form
π€π = 2π(π΄ + π΅π) + π2 Β· 2πβ1
= 2π[οΈπ΄ + π΅π +
π2
2
]οΈ, π β₯ 0 .
Ioan Despi β AMTH140 11 of 12
Examples
Hence we have πΆ = 12 and π£π = π22πβ1.
The general solution π€π of the nonhomogeneous recurrence relation isthus π€π = π’π + π£π, and hence takes the form
π€π = 2π(π΄ + π΅π) + π2 Β· 2πβ1
= 2π[οΈπ΄ + π΅π +
π2
2
]οΈ, π β₯ 0 .
Ioan Despi β AMTH140 11 of 12