5 Linear diﬀerence equations - Examples of Sequences - eBooks and textbooks from bookboon.com

Calculus 3c-1

48 Example 5.3 Find that solution of the diﬀerence equation

x_k+x_k−1=k, k≥1, for which x₀= 2.

A. Linear, inhomogeneous diﬀerence equation of ﬁrst order.

D. Guess a particular solution.

I. If we guess onx_k=αk+β, then we get by insertion that αk+β+α(k−1) +β = 2αk= (2β−α),

which is equal to the variablekforα=1

2 andβ= 1 4.

Since the characteristic polynomialR+ 1 has the rootR=−1, the complete solution is given by x_k= 1

2k+1

4+c·(−1)^k, k∈N₀, c∈R. Fork= 0 we get

c=x₀−1

4 = 2−1 4 = 7

4, thus the wanted solution is

x_k= 1 2k+1

4+7

4 ·(−1)^k, k∈N0.

Example 5.4 Find that solution of the diﬀerence equation x_k−x_k−1=k, k≥1,

for which x₂= 4.

A. Linear, inhomogeneous diﬀerence equation of ﬁrst order.

D. Solve the corresponding homogeneous equation. Then guess a particular solution.

I. The characteristic polynomial R−1 has the root R = 1. Thus, the complete solution of the homogeneous equation is given by

x_k=c, k∈N0, c∈R.

Since alreadyx_k= 1 is a solution of the homogeneous equation and sincek·1 occurs on the right hand side of the equation, we guess in analogy with the method of solving diﬀerential equations on a solution of the structure

x_k=αk²+βk, k∈N0.

Calculus 3c-1

We get by insertion, x_k−x_k−1=α&

k²−(k−1)²'

+β{k−(k−1)}=α(2k−1) +β= 2αk+ (β−α).

This expression is equal to the variablek, ifα=β= 1

2, thus the complete solution is x_k= 1

2k²+1

2+c, k∈N₀, c∈R.

Ifk= 2, we get the condition 4 = 1

2 ·2²+1

2·2 +c= 3 +c, from whichc= 1, and the solution is

x_k= 1 2k²+1

2k+ 1, k∈N₀.

Example 5.5 At some given time we have 1000 bacteria in a culture of bacteria . We assume in general that the number of bacteria is increased by 250 % every second hour. How many bacteria will there be after 24 hours?

A. Exponential growth (diﬀerence equation). Notice that since we are looking at 12·2 = 24 hours, we shall ﬁndx₁₂.

D. There are some problems here with the interpretation of the text. If theincrease really is 250 %, then the corresponding diﬀerence equation becomes

x_k=x_k−1+5

2x_k−1= 7 2x_k−1.

If the meaning instead is that the increase is 250 % of the previous value at time (k−1)2, then the diﬀerence equation becomes

x_k= 5 2x_k−1.

Since there is some linguistic uncertainty in the original text (I do not remember where I found this example), we shall as an exercise go through the solving of both equations.

I. We havex₀= 1000.

1) In the ﬁrst interpretation we get x₁₂=

7 2

₁₂

·1000≈3 379 220 508.

2) In the second interpretation we get x₁₂=

5 1

₁₂

·1000≈59 604 645.

Linear difference equations

Example 5.6 A man pays every quarter 600 euro into his account. The interest is 11 % p.a., where the interest is added every quarter. When will there be 25 000 euro on the account?

A. Diﬀerence equation for savings.

D. Write down the model of diﬀerence equation and then solve this equation.

I. Let x_k denote the capital at the k-th quarter after his payment. Then x₀ = 600, and since the interest per quarter is 11

4 %, we get the equation x_k= 600 +

1 + 11

400

x_k−1, k∈N. This is then rewritten as the diﬀerence equation

x_k−411

400x_k−1= 600.

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Calculus 3c-1

The solution of the corresponding equation is x_k =c·

411 400

, k∈N0, c∈R.

Then we guess a particular solution of the structurex_k=α,k∈N0. We get by insertion x_k−411

400x_k−1=− 11 400α, which is equal to 600 for

α=−240 000 11 .

Thus the complete solution is x_k =−240 000

11 +c· 411

400 _k

, k∈N0, c∈R. Fork= 0 we get

c= 600 + 240 000

11 =246 600 11 , so the solution becomes

x_k =−240 000

11 +246 600 11 ·

411 400

, k∈N0.

Finally, we shall ﬁnd the smallestk∈N0, for which x_k ≥25 000. Hence we shall ﬁnd the smallest k∈N0, for which

246 600 11 ·

411 400

≥25 000 +240 000

11 =515 000 11 . This is rearranged as

411 400

≥ 515 000

246 000 = 2575 1233, from which

k≥ ln

2575 1233

ln 411

400

≈ 0,7364

0,0271 ≈27,15.

Hence, we see that after 28 quarters, corresponding to 7 years, we get x_k ≥ 25 000 for the ﬁrst time.

C. Weak control. There is paid 4·600 = 2400 euro per year, hence 16 800 euro in 7 years, so the result looks reasonable. considering the high (and today unrealistic) interest.

Linear difference equations

Example 5.7 In the following examples we shall deal with annuity loans. The background is in general the following:

A loan on G₀ euro is repaid with a ﬁxed payment of Aeuro per settling period, and the interest to be paid of the debt is r per settling period (where we give r as a usual fraction, and not in %). LetG_n denote the remaining debt after nsettling periods.

1) Prove thatG_n satisﬁes

G_n−(1 +r)G_n−1=−A, n≥1, and then ﬁndG_n.

2) Establish the condition that the debt is repaid.

A. Annuity loans.

D. Analyze the situation at the end of then-th settling period in order to find the difference equation of the problem. Then solve this difference equation.

I. 1) The remaining debtG_n at then-th settling period is equal to the remaining debtG_n−1at the (n−1)-th settling period, plus the interest,×G_n−1, and minus the paymentA, i.e.

G_n = (1 +r)G_n−1−A,

which we rearrange as the solution G_n−(1 +r)G_n−1=−A, n≥1.

Here we guess a particular solution of the form G_n=α. We get by insertion (i.e. testing this solution) that

G_n−(1 +r)G_n−1=α−(1 +r)α=−rα=−A, and a particular solution is the constant sequence

G_n =α= A

r, n∈N0.

Since the corresponding homogeneous equation has the solutionc·(1+r)ⁿ, the complete solution of the inhomogeneous equation is given by

G_n =A

r +c·(1 +r)ⁿ, n∈N0, c∈R. Forn= 0 we get the condition

G₀= A

r +c, i.e. c=G₀−A r, and the wanted solution becomes

G_n =A r +

G₀−A

·(1 +r)ⁿ, n∈N₀.

Calculus 3c-1

2) The debt will be paid back, if and only if G_n at some time becomes ≤0. This means that G₀−A

r <0, hence A > r·G₀,

which is only expressing the reasonable fact that the payment must be bigger than the interest in one settling period of the original loan.

Example 5.8 LetG₀ andA be given, and assume thatA is suﬃciently large to assure that the loan will be paid. Find a formula for the number of settling periods which is needed to pay the debt (thus the smallest number nof settling periods for which G_n≤0.)

A. A continuation of Example 5.7.

D. Apply the solution of Example 5.7.

I. According to Example 5.7 the remaining debtG_n is given by G_n= A

r +

G₀−A r

·(1 +r)ⁿ, n∈N₀,

where we must assume thatG₀−A

r <0, i.e.A > r·G₀.

We shall ﬁnd the smallestn∈N₀, for whichG_n≤0, i.e. the smallestn∈N₀, for which A

r −G₀

·(1 +r)ⁿ≥ A

r, i.e. (1 +r)ⁿ≥ A A−r·G₀. We get by taking the logarithm that

n≥ lnA−ln(A−r·G₀) ln(1 +r) .

Example 5.9 Let the number of settling periods be ﬁxed to N. Find a formula for the (constant) payment, by which the debt is paid in precisely N settling periods.

A. A continuation of Example 5.8.

D. Apply the solution of Example 5.8.

I. IfA > r·G₀, we see from Example 5.8 that we shall ﬁndA, such that

N = lnA−ln(A−r·G₀) ln(1 +r) =−

A−r·G₀ A

ln(1 +r) =−

1−rG₀ A

ln(1 +r) , which we rewrite as

1−rG₀ A

=−N ln(1 +r) = ln&

(1 +r)^−N' .

Linear difference equations

54 We get from here the condition

(1 +r)^−N = 1−rG₀

A , i.e. rG₀

A = 1−(1 +r)^−N, hence

A=G₀· r

1−(1 +r)^−N =G₀· r(1 +r)^N

(1 +r)^N−1 =G₀·(1 +r)^N+1−(1 +r)^N (1 +r)^N−1 .

Example 5.10 Assume that G₀ is 100 000 euro and that the annual interest is 9 % and that there are 4 settling periods (quarters) per year. When is the loan repaid if the payment is 3 000 euro per quarter? And when is the loan repaid, if we double the payment?

A. A continuation of the Examples 5.7–5.9.

D. Apply the results from Examples 5.7–5.9.

I. First calculate

G₀= 100 000, r= 9

400 and A= 3 000.

Calculus 3c-1

We get rG₀= 9

400·100 000 = 9·250 = 2 250<3 000 =A,

which guarantees that the loan will be repaid with 3 000 euro in payment per settling period, cf.

Example 5.7.

It follows from Example 5.9 that n≥ lnA−ln(A−rG₀)

ln(1 +r) = ln 3000−ln 750 ln

1 + 9

400

= ln 4 ln409

400

≈62,3,

thus the loan is repaid after 63 settling periods, corresponding to 15³₄ years.

WhenA is doubled to 6 000 euro, we getA−rG₀= 3 750, hence

n≥ lnA−ln(A−rG₀) ln(1 +r) =

ln6000 3750 ln409

400

≈21,12,

and the loan is repaid after (only) 22 settling periods, corresponding to 5¹₂˚ar.

Example 5.11 Assume that G₀ is 100 000 euro and that the annual interest is 9 % and that the payments are quarterly. How big shall we choose the payment, if the loan is repaid after 20 and 30 years, respectively?

A. A continuation of the Examples 5.7-5.10.

D. Use the previous results from Example 5.9.

I. Here G₀ = 100 000 and r = 9

400. In the ﬁrst case, N = 4·20 = 80, and in the second case is N = 4·30 = 120. It follows from Example 5.9 that

A=rG₀(1 +r)^N

(1 +r)^N−1 = rG₀ 1−(1 +r)^−N. 1) IfN = 80, then

A= 9

400 ·100 000 1−

400 409

₈₀ = 2250 1−

400 409

₈₀ = 2 706,38 euro.

2) N˚ar N= 120, er

A= 2250

1− 400

409

₁₂₀ = 2 417,40 euro

Linear difference equations

56 Example 5.12 Find that solution of the diﬀerence equation

x_k+ 2x_k−1= cos π

, k≥1, for which x₀= 0.

A Linear, inhomogeneous diﬀerence equation of ﬁrst order.

D. Start by ﬁnding the solution of the homogeneous equation. Then analyze the right hand side (guess a complex solution).

I. The characteristic polynomial R+ 2 has the rootR =−2, so the complete solution of the homogeneous equation is given by

x_k=c·(−2)^k, k∈N₀, c∈R.

Now, cos π

2 k

= Re&

i^k'

, so if we insertx_k =α i^k, we get x_k+ 2x_k−1=α i^k+ 2α i^k−1=α(1−2i)i^k,

which is equal toi^k, if α= 1

1−2i =1 + 2i 5 . Thus, a particular solution is

x_k= Re

1 + 2i 5 i^k

=1 5Re&

i^k+ 2i^k+1'

= 1 5 cos

k·π

+2 5 cos

(k+ 1)π 2

, and the complete solution becomes

x_k= 1 5 cos

k·π

+2 5 cos

(k+ 1)π 2

+c·(−2)^k. It follows from the initial condition that

x₀= 0 = 1

5+ 0 +c, i.e. c=−1 5. The wanted solution becomes

x_k = 1 5

(−2)^k+ cos π

2 k

+ 2 cos π

2 (k+ 1)

= 1

(−2)^k+ cos π

2 k

−2 sin π

, k∈N₀.

Calculus 3c-1

Example 5.13 Let a_n = 1 + 2 +· · ·+n, n≥1. Find a difference equation which is fulfilled bya_n. Then find a formula fora_n.

A. Establishment and solution of a diﬀerence equation.

D. Look ata_n−a_n−1. I. Obviously,

a_n−a_n−1=n, n≥1.

The corresponding homogeneous equation has the constant solution a_n=c, n∈N₀, c∈R.

Then we guess the structurea_n=αn²+βn. By insertion, a_n−a_n−1 = α&

n²−(n−1)²'

+β{n−(n−1)}

= α(2n−1) +β= 2αn+ (β−α) =n, thusα= 1

2 andβ =α= 1

2. The complete solution becomes a_n= 1

2n²+1

2n+c= 1

2n(n+ 1) +c, n∈N₀, c∈R. Sincea₁= 1 = 1

2·1·2 +c= 1 +c, we havec= 0, and the searched solution is a_n= 1

2n(n+ 1), n∈N₀.

Example 5.14 Let a_n= 1 + 2²+· · ·+n²,n≥1. Find a difference equation which is fulfilled bya_n. Then find a formula fora_n.

A. Establish a (simple) diﬀerence equation.

D. Find such a diﬀerence equation fora_n, and solve it.

I. It is immediately seen that a_n−a_n−1=n².

The corresponding homogeneous equation has the constant sequencea_n=c,c∈R, as a solution.

Then we guess a particular solution of the form a_n=αn³+βn²+γn,

hence by insertion, a_n−a_n−1 = α&

n³−(n−)³' +β&

n²−(n−1)²'

+γ{n−(n−1)}

= α&

3n²−3n+ 1'

+β{2n−1}+γ

= 3αn²+ (−3α+ 2β)n+ (α−β+γ).

Linear difference equations

58 This expression is equal ton², if and only if

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

3α= 1,

−3α+ 2β= 0, α−β+γ= 0,

i.e.

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

α=¹₃, β= ¹₂,

γ=β−α=¹₆. A particular solution is then

a_n=1 3n³+1

2n²+1 6n= 1

6n(2n²+ 3n+ 1) = 1

6n(2n+ 1)(n+ 1).

Sincea₁= 1

6·1·(2 + 3 + 1) = 1, we see that this is in fact the wanted solution, so a_n=1

6n(n+ 1)(2n+ 1), n∈N. The complete solution is of course

a_n=1

6n(n+ 1)(2n+ 1) +c, n∈N, c∈R.

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Calculus 3c-1

Example 5.15 Find the complete solution of the following diﬀerence equations (1) x_k−5x_k−1+ 6x_k−2= 0, k≥2,

(2) x_k−6x_k−1+ 9x_k−2= 0, k≥2, (3) x_k+ 2x_k−1+ 2x_k−2= 0, k≥2.

A. Linear homogeneous diﬀerence equations of second order.

D. Find the roots of the characteristic polynomials and apply the solution formula.

I. 1) The characteristic polynomial R²−5R+ 6 has the simple rootsR = 2 andR= 3, hence the complete solution is

x_k =c₁·2^k+c₂·3^k, k∈N0, c₁, c₂∈R.

2) The characteristic polynomial R²−6R+ 9 has the double root R = 3, hence the complete solution is

x_k =c₁·3^k+c₂·k·3^k, k∈N₀, c₁, c₂∈R.

3) The characteristic polynomial R²+ 2R+ 2 has the two complex conjugated roots R=−1±i=√

2 exp

±i3π 4

hence the complete solution is given by x_k =c₁(√

2)^kcos 3π

4 k

+c₂(√ 2)^ksin

3π 4 k

, k∈N₀, wherec₁,c₂∈Rare arbitrary constants.

Example 5.16 Find in each of the following cases that solution of the diﬀerence equation which also satisﬁes the given initial condition.

(1) x_k−7x_k−1+ 10x_k−2= 0, k≥2, x₀= 3, x₁= 15.

(2) 9x_k+ 12x_k−1+ 4x_k−2= 0, k≥2, x₀= 1, x₁= 4.

(3) x_k+ 4x_k−2= 0, k≥2, x₀=x₁= 1.

A. Linear homogeneous diﬀerence equations of second order with given initial conditions.

D. Find the roots of the characteristic polynomials and apply some convenient solution formula. Then insert into the initial conditions.

I. 1) The characteristic polynomialR−7R+ 10 has the two simple roots R= 2 and R= 5, hence the complete solution is

x_k =c₁·2^k+c₂·5^k, k∈N0, c₁, c₂∈R.

Linear difference equations

60 It follows from the initial conditions that

x₀= 3 =c₁+c₂,

x₁= 15 = 2c₁+ 5c₂, i.e.

2c₁+ 2c₂= 6, 2c₁+ 5c₂= 15,

thus 3c₂= 15−6 = 9, i.e.c₂= 3 andc₁= 0. The wanted solution is x_k = 3·5^k, k∈N₀.

2) The characteristic polynomial 9R+ 12R+ 4 = (3R+ 2)² has the double rootR=−2 3, hence the complete solution is

x_k =c₁

−2 3

+c₂·k

−2 3

, k∈N₀, c₁, c₂∈R.

It follows from the initial conditions that x₀= 1 =c₁ and x₁= 4 =−2

3(c₁+c₂),

thusc₁= 1 and c₁+c₂=−6, i.e.c₂=−7. The wanted solution is x_k = (1−7k)·

−2 3

, k∈N₀.

3) The characteristic polynomial R²+ 4 has the two complex conjugated roots R=±2i= 2 exp

±iπ 2

. Hence, the complete solution is

x_k =c₁·2^kcos π

+c₂·2^ksin π

2 k

, k∈N0, wherec₁,c₂∈Rare arbitrary constants.

It follows from the initial conditions that

x₀= 1 =c₁ and x₁= 1 = 2c₂, i.e. c₂= 1 2. Thus, the wanted solution becomes

x_k = 2^k

cosπ 2 k

+1 2 sinπ

2 k

, k∈N₀.

Calculus 3c-1

Example 5.17 Find the complete solution of the diﬀerence equation x_k+ 4x_k−1+ 4x_x−2= 7, k≥2.

A. Linear inhomogeneous diﬀerence equation of second order.

D. Find the roots of the characteristic polynomial and apply the solution formula when solving the homogeneous equation. Finally, guess the structure of a particular solution and apply the linearity.

I. We guess a particular solution as a constant sequence,x_k =c. It is seen by insertion thatx_k= 7 9, k∈N₀, is a particular solution.

The characteristic polynomialR²+ 4R+ 4 = (R+ 2)² has the double root R = −2, hence the complete solution becomes

x_k= 7

9+c₁(−2)^k+c₂k(−2)^k, k∈N0, c₁, c₂∈R.

Example 5.18 Find that solution of the diﬀerence equation x_k+ 2x_k−1+ 2x_k−2= 5^k, k≥2,

for which x₀=x₁=1 5.

A. Linear inhomogeneous diﬀerence equation of second order with initial conditions.

D. Find the roots of the characteristic polynomial and apply a solution formula for the homogeneous equation. Then guess the structure of a particular solution and exploit the linearity. Insert ﬁnally into the initial conditions and ﬁnd the constants.

I. The characteristic polynomialR²+ 2R+ 2 has the two complex conjugated roots R=−1±i=√

2 exp

±3π 4 i

The complete solution of the homogeneous equation is x_k=c₁(√

2)^kcos 3π

4 k

+c₂(√ 2)^ksin

3π 4 k

, k∈N0, wherec₁,c₂∈Rare arbitrary constants.

Then we guess a particular solution of the structurex_k =α·5^k. We get by insertion x_k+ 2x_k−1+ 2x_k−2=α

5^k+ 2·5^k−1+ 2·5^k−2

= α

25(25 + 10 + 2)·5^k= 37 25α·5^k. This expression is equal to 5^k, ifα= 25

37, thus the complete solution becomes x_k= 25

375^k+c₁(√ 2)^kcos

3π 4 k

+c₂(√

2)^ksin 3π

4 k

, k∈N0,

Linear difference equations

62 wherec₁ andc₂∈Rare arbitrary constants.

It follows from the initial conditions that x₀= 1

5 = 25 37+c₁, hence

c₁= 1 5−25

37 =37−125

185 =−88 185, and

x₁= 1 5 = 125

37 −c₁

√2

√2 +c₂

√2

√2 =125

37 −c₁+c₂, whence

c₂= 1 5+1

5 −25 37−125

37 = 2 5 −150

37 =74−750

185 =−676 185. The wanted solution is therefore

x_k =25

37·5^k− 88 185(√

2)^kcos 3π

4 k

−676 185(√

2)^ksin 3π

4 k

, k∈N0.

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Calculus 3c-1

Example 5.19 Find that solution of the diﬀerence equation x_k+ 3x_k−1+ 2x_k−2= 3^k, k≥2,

for which x₀= 0andx₁= 0.

A. Linear inhomogeneous diﬀerence equation of second order with initial conditions.

D. Find the roots of the characteristic polynomial and apply a solution formula for the complete solution of the homogeneous equation. Then guess the structure of a particular solution and exploit the linearity. Insert finally into the final conditions and find the constants.

I. The characteristic polynomial R²+ 3R+ 2 has the two simple rootsR=−1 andR=−2, hence the complete solution of the homogeneous diﬀerence equation becomes

x_k=c₁(−1)^k+c₂(−2)^k, k∈N0, c₁, c₂∈R.

Then we guess on a particular solution of the structurex_k=α·3^k. By insertion into the equation we get

α&

3^k+ 3^k+ 2·3^k−2'

=α

2 + 2 9

3^k =20 9 α·3^k, which is equal to 3^k forα= 9

20. Thus the complete solution is x_k= 9

20·3^k+c₁(−1)^k+c₂(−2)^k, k∈N₀, c₁, c₂∈R.

Then we get by the ﬁnal conditions,

⎧⎪

⎪⎨

⎪⎪

⎩

x₀= 0 = 9

20+c₁+c₂, x₁= 0 = 27

20−c₁−2c₂, whence

c₂=27 20+ 9

20 =36 20 = 9

5, and

c₁=−9

20−c₁=−9 20−9

5 =−45 20 =−9

4. The wanted solution is therefore

x_k= 9

20·3^k−9

4(1−)^k+9

5(−2)^k, k∈N0.

Linear difference equations

Example 5.20 Find the complete solution of the diﬀerence equations (1) x_k−x_k−2= sin

k·π

, k≥2, (2) x_k−x_k−2= cos

k·π

, k≥2.

A. Linear inhomogeneous diﬀerence equations of second order.

D. Find the roots of the characteristic equation and apply a solution formula for the solution of the homogeneous diﬀerence equation. Then try to make a complex guess of a particular solution.

I. Since sin

k·π 2

= Im i^k

, and cos

k·π 2

= Re i^k

, we can solve both problems at the same time, until we at last are forced to split into the real and the imaginary part.

The characteristic polynomialR²−1 has the two simple rootsR=±1, hence the complete solution of the homogeneous equation becomes

x_k =c₁+c₂(−1)^k, k∈N₀, c₁, c₂∈R.

Next insertx_k=α i^k. We get x_k−x_k−2=α&

i^k−i^k−2'

= 2α i^k, which is equal to i^k forα=1

2. 1) The complete solution of

x_k−x_k−2= sin

k·π 2

= Im i^k is

x_k= Im 1

2i^k

+c₁+c₂(−1)^k= 1 2 sin

k·π

+c₁+c₂(−1)^k, k∈N0, c₁, c₂∈R. 2) The complete solution of

x_k−x_k−2= cos

k· π 2

= Re i^k is

x_k= Re 1

2i^k

+c₁+c₂(−1)^k= 1 2 cos

k·π

+c₁+c₂(−1)^k, k∈N0, c₁, c₂∈R.

Calculus 3c-1

Example 5.21 Find that solution of the diﬀerence equation x_k−6x_k−1+ 9x_k−2= 3·2^k+ 17·4^k, k≥2,

for which x₀= 40andx₁= 400.

A. Linear inhomogeneous diﬀerence equation of second order with initial conditions.

D. Solve the characteristic equation; guess a particular solution.

I. The characteristic polynomial R²−6R+ 9 = (R−3)² has the double root R = 3, hence the homogeneous equation has the complete solution

x_k=c₁3^k+c₂k·3^k, k∈N0, c₁, c₂∈R.

By guessing the structure x_k = a·2^k +b·4^k, we get by insertion (i.e. checking this possible solution)

x_k−6x_k−1+ 9x_k−2=a·2^k+b·4⁴−6a·2^k−1−6b·4^k−1+ 9a·2^k−2+ 9b·4^k−2

1−3 +9 4

2^k+b

1−3

2+ 9 16

4^k= a

4 ·2^k+ b 16·4^k.

This expression is equal to 3·2^k+ 17·4⁴, if a= 12 and b = 16·17 = 272, hence the complete solution becomes

x_k= 12·2^k+ 272·4^k+c₁·3^k+c₂k·3^k, k∈N0, c₁, c₂∈R. By insertion into the conditions we get

x₀= 40 = 12 + 272 +c₁= 284 +c₁, thusc₁=−244, and

x₁= 400 = 24 + 1088 + 3c₁+ 3c₂= 1112−732 + 3c₂= 380 + 3c₂, i.e.c₂= 20

3 . The solution is

x_k= 12·2^k+ 272·4^k−244·3^k+20

3 k·3^k, k∈N0.

Linear difference equations

Example 5.22 Find the complete solution of the diﬀerence equations (1) x_k−2x_k−1+x_k−2−2x_k−3= 0, k≥3,

(2) x_k−2x_k−1−x_k−2+ 2x_k−3= 0, k≥3, (3) x_k−x_k−4= 0, k≥4.

A. Two linear homogeneous diﬀerence equations of third order, and an homogeneous diﬀerence equation of fourth order.

D. Find the roots of the characteristic polynomials and then apply a solution formula.

I. 1) The characteristic polynomial

R³−2R²+R−2 = (R−2)(R²+ 1) has the simple rootsR= 2 andR=±i= exp

±iπ 2

, thus all (real) solutions are x_k=c₁·2^k+c₂ cos

k·π

+c₃ sin

k·π 2

, k∈N0, c₁, c₂, c₃∈R.

Calculus 3c-1

2) The characteristic polynomial

R³−2R²−R+ 2 = (R−2)(R²−1) = (R−2)(R−1)(R+ 1)

has the three simple rootsR= 2 andR=±1, hence the complete solution is x_k =c₁·2^k+c₂+c₃(−1)^k, k∈N0, c₁, c₂, c₃∈R.

3) The characteristic polynomial

R⁴−1 = (R−1)(R+ 1)(R−i)(R+i)

has the four simple rootsR=±1 andR=±i, hence all (real) solutions are x_k =c₁0c₂(−1)^k+c₃cos

k·π

+c₄ sin

k·π 2

, k∈N₀, wherec₁,c₂,c₃,c₄∈Rare arbitrary constants.

Example 5.23 Find that solution of the diﬀerence equation x_k−x_k−1+ 4x_k−2−4x_k−3= 0, k≥3,

for which x₀=−1,x₂= 2,x₄= 4.

A. Linear homogeneous diﬀerence equation of third order with three (non-successive) conditions.

D. Find the roots of the characteristic polynomial and apply the solution formula. Finally, insert into the conditions.

I. The characteristic polynomial

R³−R²+ 4R−4 = (R−1)(R²+ 4)

has the three simple rootsR= 1 and R=±2i. Hence, the complete solution is x_k=c₁+c₂2^kcos

k· π

+c₃2^ksin

k·π 2

. Then by insertion into the conditions,

⎧⎨

⎩

x₀ = −1 = c₁ + c₂,

x₁ = 2 = c₁ + 2c₃,

x₄ = 4 = c₁ + 16c₂, from which we get 15c₂= 5, thusc₂=1

3, and c₁=−4

3. Then ﬁnally 2c₃= 2 +4

3, i.e. c₃= 1 + 2 3 = 5

3. The wanted solution is

x_k=−4 3 +1

3·2^kcos

k·π 2

+5 3 ·2^ksin

k·π

, k∈N₀.

Linear difference equations

Example 5.24 Find the complete solution of the diﬀerence equation x_k−2x_k−1+x_k−2−2x_x−3= 4, k≥3.

A. Linear inhomogeneous diﬀerence equation of fourth order. The corresponding homogeneous difference equation is treated in Example 5.22 (1).

D. Find the roots of the characteristic polynomial; then guess a solution of the inhomogeneous equation.

I. The characteristic polynomial

R³−2R²+R−2 = (R−2)(R²+ 1)

has the three simple roots R= 2 and R=±i. It is seen by inspection thatx_k =−2,k∈N0, is a particular solution. Hence the complete solution is

x_k =−2 +c₁·2^k+c₂ cos

k·π 2

+c₃ sin

k·π 2

, k∈N₀, wherec₁,c₂,c₃∈Rare arbitrary constants.

Example 5.25 Find the complete solution of the diﬀerence equation x_k−2x_k−1−x_k−2+ 2x_k−3=−2, k≥3.

A. Linear inhomogeneous diﬀerence equation of third order. The corresponding homogeneous equation was dealt with in Example 5.22 (2).

D. Find the roots of the characteristic polynomial; then guess a solution of the inhomogeneous equation.

I. The characteristic polynomial

R³−2r²−R+ 2 = (R−2)(R−1)(R+ 1)

has the three simple rootsR= 2 andR=±1. Since alreadyR= 1 is a root, corresponding to the solution x_k =c,k∈N0, of the homogeneous equation, our guess must be modiﬁed, so we guess a particular solution of the form x_k =a·k. We get by insertion,

x_k−2x_k−1−x_k−2+ 2x_k−3=a{k−2(k−1)−(k−2) + 2(k−3)}

=a{k−2k+ 2−k+ 2 + 2k−6}=−2a,

which is equal to −2 for a= 1. Hence, the complete solution is given by x_k =k+c₁+c₂(−1)^k+c₃·2^k, k∈N0, c₁, c₂, c₃∈R.

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