LAPLACE TRANSFORMS
BORIS HASSELBLATT
CONTENTS
1. Complex numbers, Euler’s formula 1
2. Homogeneous differential equations 6
3. Undetermined coefficients 8
4. Laplace transforms 10
5. Partial fractions 11
References 16
The purpose of these notes is to introduce complex numbers and their use in solving ordinary differential equations. The main feature is that trigonometric functions can be omitted from the methods even when they arise in a given problem or its solution.
One one hand this approach is illustrated with the method of undeter-mined coefficients, where this approach offers a parallel development to the one in the text [1]. On the other hand, this is done with the Laplace transform, where the use of complex numbers replaces a great deal of machinery needed otherwise and at the same time covers situations that the methods in the text do not. The main effort here is to show how to produce partial-fractions decompositions using complex numbers.
The reason the method of undetermined coefficients is revisited here in the complex context is that fluency with this method is very helpful in using the Laplace transform method reliably.
The exercises included here are not for credit but will help you read this text actively.
1. COMPLEX NUMBERS, EULER’S FORMULA
We introduce the symbol i with the property
i2= −1
A complex number is an expression that can be written in the form a +i b with real numbers a and b.
1.1. Real and imaginary part, complex conjugate. If a and b are real numbers, then a is called the real part of a + i b, and b is called the
imgi-nary part. (Note that the imagiimgi-nary part is a real number!)
The expression a − i b is called the complex conjugate of a + i b. It is sometimes denoted by a bar:
a + i b = a − i b.
Adding a complex number and its complex conjugate always gives a real number:
a + i b + a − i b = 2a.
This is twice the real part. So, if we are given a complex number z = a +i b in any form, we can express the real part as
ℜ(z) = real part of z =z + z 2 . The imaginary part can be expressed as (check!)
ℑ(z) = imaginary part of z =z − z 2i . 1. Exercise. Verify that z = z.
2. Exercise. Verify that any real number x satisfies x = x.
3. Exercise. Verify that a complex number z satisfying z = z is a real num-ber.
1.2. Multiplying complex numbers. To multiply two complex numbers just use i2= −1 and group terms:
(a + i b)(c + i d) = ac + ai d + i bc + i bi d = ac − bd + i (ad + bc). Multiplying a complex number and its complex conjugate always gives a real number:
(a + i b)(a − i b) = a2+ b2.
We callpa2+ b2the absolute value or modulus of a + i b:
|a + i b| =pa2+ b2
4. Exercise. Verify that |z| =pzz.
5. Exercise. Verify that z1z2= z1z2.
1.3. Dividing complex numbers. To divide two complex numbers and write the result as real part plus i ×imaginary part, multiply top and bot-tom of this fraction by the complex conjugate of the denominator:
a + i b c + i d = a + i b c + i d × c − i d c − i d = (a + i b)(c − i d) c2+ d2 = ac + bd + i (bc − ad) c2+ d2 .
1.4. Factoring polynomials. Factoring polynomials is no harder (or eas-ier) when complex numbers are allowed but in this case all factors are linear. The reason is that factors x − α are now legal even when α is com-plex.
7. Example. The polynomial s2+ 1 is irreducible over the real numbers, but we have
s2+ 1 = (s − i )(s + i ).
1.5. Euler’s formula. Complex numbers are useful in our context because they give Euler’s formula
eiθ= cos θ + i sin θ.
This formula is easy to remember. In case you are not sure whether to attach the i to the cos or to the sin, just plug inθ = 0.
It’s worthwhile recalling some power series to see why this is so. We have ez= 1 + z + z2/2 + ... = ∞ X n=0 zn n! 1 cos z = 1 − z2/2 + ... = ∞ X n=0 (−1)nz2n (2n)! sin z = z − z3/6 + ... = ∞ X n=0 (−1)nz2n+1 (2n + 1)! With z = i θ this gives
eiθ= ∞ X n=0 (iθ)n n! = ∞ X n=0 inθn n! = ∞ X n=0 i2nθ2n (2n)! + ∞ X n=0 i2n+1θ2n+1 (2n + 1)!
Here we have simply summed even and odd powers separately. The rea-son we did this is that we can rewrite
i2n= (i2)n= (−1)n and i2n+1= i (i2)n= i (−1)n.
If we pull out the leading i s from the sum over odd powers of t , we find
eiθ= ∞ X n=0 (−1)nθ2n (2n)! + i ∞ X n=0 (−1)nθ2n+1 (2n + 1)! = cos θ + i sin θ.
In summary, we will use the “forward” and “backward” Euler formulas eiθ= cos θ + i sin θ cosθ =e iθ+ e−i θ 2 sinθ = eiθ− e−i θ 2i = i 2(e −i θ− eiθ)
8. Exercise. Show that ez= ez.
9. Exercise. Verify that ifλ is complex and aeλt+beλtis real for all t , then
b = a via the following steps.
(1) Rewrite the given information using Exercise 2 (and Exercise 8). (2) The deriviative of aeλt+beλtis also real—compute it and proceed
as in the previous part.
(3) Take t = 0 in the expressions obtained in both previous parts. 1.6. The complex plane. While real numbers can be geometrically rep-resented by a number line, complex numbers can be reprep-resented by a plane, called the complex plane. Taking the real number line as a hori-zontal axis, one can introduce a vertical axis with i as the unit on it. Then
a +i b can be understood as representing a point in this plane in terms of
cartesian coordinates.
The power of this representation is related to the fact that Euler’s for-mula represents the same plane but using polar coordinates. (Recall that polar coordinates in the plane are of the form (r cosθ,r sinθ), where r is the distance from the origin andθ the angle with the horizontal axis.) To see this, we contemplate
z = ea+i b= eaei b= eacos b + i easin b. Here eaplays the role of r in polar coordinates, and indeed,
|ea+i b| = |z| =pz ¯z =pea+i bea−i b=pe2a= ea.
So in the complex plane, ea+i b has distance ea from the origin and lies in a direction relative to the horizontal axis given by the angle b, which is called the argument of ea+i b.
One consequence is that multiplication of complex numbers can be interpreted geometrically:
eaei b· ecei d= ea+i b· ec+i d= eaecei (b+d).
The absolute value of this product is eaec, which is the product of the absolute values of the 2 numbers. The argument is the sum of the argu-ments of the 2 numbers. Another way of saying this, since the argument is an angle, is that multiplying a complex number by another rotates and scales it.
1.7. Exponential shift. As in [1, p. 123], the exponential shift works for complex exponentials (you can check that the calculation on the bottom half of that page do not use thatλ is real):
P (D)[eλty] = eλtP (D + λ)y.
1.8. Hyperbolic functions. While we think of t as a real variable in Eu-ler’s formula, one gets an interesting result when one plugs imaginary numbers into cos and sin:
cos i x =e i (i x)+ e−i (i x) 2 = e−x+ ex 2 = ex+ e−x 2 =: cosh x.
Here, cosh x is the hyperbolic cosine function defined by this last equality. Likewise, sin i x =e i (i x) − e−i (i x) 2i = e−x− ex 2i = i ex− e−x 2 = i sinh x.
1.9. The trigonometric-identity machine. Euler’s formula produces many trigonometric identities by using the rules of exponents.
10. Example. Find formulas for cos(a + b) and for sin(a + b).
We start with the law of exponents ei (a+b) = ei aei b and rewrite every exponential using Euler’s formula. On the left we have
ei (a+b)= cos(a + b) + i sin(a + b). On the right we have
ei aei b=¡ cos a + i sin a¢¡cosb + i sinb¢
=¡ cos a cosb − sin a sinb¢ + i¡sin a cosb + cos a sinb¢. Since these two things are the same, we get
cos(a+b)+i sin(a+b) =¡ cos a cosb−sin a sinb¢+i¡sin a cosb+cos a sinb¢. Matching real and imaginary parts, we get
cos(a +b) = cos a cosb −sin a sinb and sin(a +b) = sin a cosb +cos a sinb. 11. Example. To get double-angle formulas, take a = b in the previous example. Note that cos 2a then simplifies to cos2a − sin2a.
12. Example. To obtain triple-angle formulas use the law e3i a =¡ei a¢3 and rewrite every term using Euler’s formula, multiply out and sort terms. (Or use Example 10 with b = 2a and use double-angle formulas along the way.)
13. Example. Should you ever need a formula for cos(a + b + c), you can start with the law of exponents ei (a+b+c)= ei aei bei cand rewrite every ex-ponential using Euler’s formula, then multiply out and sort terms.
14. Example ([1, p. 158]). Second-order differential equations often have solutions that can be written as
x(t ) = ℜ(ceλt) =c 2e λt+c¯ 2e ¯ λt.
If we are given initial conditions x(0) = x0 and x0(0) = v0 and writeλ =
−σ + i ω, then we find that x0= x(0) = ℜ(ceλ0) = ℜc and
v0= x0(0) = ℜ(cλeλ0) = ℜ(cλ) = ℜ((x0+ i ℑc) · (−σ + i ω)) = −σx0− ωℑc,
so c = ℜc + i ℑc = x0− iv0+σxω 0
The solution can be rewritten in a way that involves only a sine or a cosine, but not both. To see how, write c = ea−i αto get
c 2e λt+c¯ 2e ¯ λt =1 2e
a−i αe(−σ+i ω)t+ea+i αe(−σ−i ω)t= eae−σtei (ωt−α)+ e−i (ωt −α)
2 .
This means that
x(t ) = ℜ(ceλt) = Ae−σtcos(ωt − α), where A = ea= |c| =
q
x20+ (v0+σx0
ω )2and
cosα + i sinα = eiα= e−i α= 1 eae a−i α= c¯ A= x0 A + i v0+ σx0 ωA .
2. HOMOGENEOUS DIFFERENTIAL EQUATIONS
If one uses complex numbers, then the method described in [1, Sec-tions 2.5, 2.6] looks slightly different, and there are a few choices one can make.
The point of [1, p. 131] is that the method of [1, Section 2.5] works even for complex roots of the characteristic polynomial, but that some care should be taken to make sure one ultimately writes down real solutions only.
15. Example ([1, Example 2.6.1]). Solve (D2+ 4D + 5)x = 0.
The characteristic polynomial has rootsλ±= −2 ± i , which gives solu-tions eλ±t, or eλ+t and eλ−t, or
e(−2+i )tand e(−2−i )t.
However, neither of these is a real solution2. There are different ways of obtaining real solutions. For the purpose of finding the general solution, the easiest is to choose, as in [1, Section 2.6] but with different notation, the solutions
ℜ(e(−2+i )t) and ℑ(e(−2+i )t).
(Think about why these are solutions, and why these 2 “contain the same information” as the pair of complex solutions above.) This gives the for-mulas at the end of [1, Section 2.6].
16. Example ([1, Exercise 2.6.17]). Solve the initial-value problem (5D2+ 2D + 1)x = 0; x(0) = 0, x0(0) = 1.
The roots of the characteristic polynomial areλ±= −15±25i , giving the
“general solution” c+eλ+t+ c −eλ−t, or c+e(−15+ 2 5i )t+ c−e(− 1 5− 2 5i )t.
This is not quite “proper” since it involves complex functions, but unlike in the previous example, this is also not the desired final answer anyway since we wish to solve an initial-value problem. While we could proceed as before and write out the general solution in real terms and then de-termine the correct coefficients from the initial values, we can instead directly do so with the complex solution above.
The condition x(0) = 0, x0(0) = 1 applied to x(t ) = c+e(− 1 5+ 2 5i )t+ c−e(− 1 5− 2 5i )t gives c++ c−= 0, c+λ++ c−λ−= 1.
Cramer’s rule gives
c+= detµ0 1 1 λ− ¶ detµ 1 1 λ+ λ− ¶ = − 1 λ−− λ+ = 1 2 ·25i = 5 2 1 2i.
Now, Exercise 9 (on page 4) or an analogous computation or the first equation tells us that c− = −c+ = −52
1
2i. Therefore the solution of the
initial-value problem is 5 2 1 2ie (−15+25i )t −5 2 1 2ie (−15−25i )t =5 2e −t /5·e 2 5i t− e− 2 5i t 2i = 5 2e −t /5sin2 5t . Which version works better is a matter of taste; the one presented here leads directly to the coefficients from the complex form with a compu-tation involving 2 equations in 2 unknowns, and it is reassuring (basic sanity check) that the end result is real.
17. Example ([1, Exercise 2.6.17] with real functions). Solve the initial-value problem
(5D2+ 2D + 1)x = 0; x(0) = 0, x0(0) = 1.
The roots of the characteristic polynomial areλ±= −15±25i , giving by
Ex-ercise 9 the general solution ceλ+t+ ¯ceλ−t, or
ce(−15+ 2 5i )t+ ¯ce(− 1 5− 2 5i )t.
Since this is a real function by Exercise 9, it is quite proper, even though the coefficients are complex. There is only one (complex) coefficient to determine. The condition x(0) = 0, x0(0) = 1 applied to x(t ) = ce(−15+ 2 5i )t+ ¯ce(− 1 5− 2 5i )t gives c + ¯c = 0, cλ++ ¯c ¯λ+= 1.
The first equation (which says that c is imaginary) gives ¯c = −c, so the
second equation gives
1 = c · (λ+− ¯λ+) = c · 2i ℑλ+= c · 2i 2 5, so c =5 2 1 2i.
As above, one then finds that the solution of the initial-value problem is
x(t ) =5
2e
−t /5sin2
5t .
3. UNDETERMINED COEFFICIENTS
The method of undetermined coefficients works much like in [1, Sec-tion 2.7], but we use complex exponentials and complex factoring of poly-nomials where appropriate.
18. Example ([1, Example 2.7.5]). Solve
(N) (D2− 4)x = 1 + 65etcos 2t .
Since D2−4 = (D −2)(D +2), the general solution of the associated homo-geneous equation
is
x = H(t) = c1e2t+ c2e−2t.
The forcing term of (N) can be rewritten as 1 +652e(1+2i )t+652e(1−2i )t and is therefore annihilated by
D(D − (1 + 2i ))(D − (1 − 2i )).
Therefore a particular solution x = p(t) of (N) will also satisfy (H*) D(D − (1 + 2i ))(D − (1 − 2i ))(D − 2)(D + 2)x = 0.
This has characteristic polynomial r (r − (1 + 2i ))(r − (1 − 2i ))(r − 2)(r + 2), so p(t ) has to be of the form
p(t ) = k1+ k2e(1+2i )t+ k3e(1−2i )t+ k4e2t+ k5e−2t.
19. Remark. When we use this method later to predict what terms to ex-pect in applying the Laplace transform method, then we can stop at this point.
We get a simplified guess by omitting the terms that solve (H):
p(t ) = k1+ k2e(1+2i )t+ k3e(1−2i )t.
To determine the (as yet undetermined) coefficients k1, k2and k3, insert
this simplified guess into (N), written with the forcing term on the left: 1 +65 2 e (1+2i )t+65 2 e (1−2i )t= (D2 − 4)p(t ) = (D2− 4)[k1+ k2e(1+2i )t+ k3e(1−2i )t] = (D2− 4)k1+ (D2− 4)k2e(1+2i )t + (D2− 4)k3e(1−2i )t
(complex exponential shift) = −4k1+ e(1+2i )t([D + (1 + 2i )]2− 4)k2
+ e(1−2i )t([D + (1 − 2i )]2− 4)k3
= −4k1+ e(1+2i )t((1 + 2i )2− 4)k2
+ e(1−2i )t((1 − 2i )2− 4)k3
Since the functions 1, e(1+2i )t and e(1−2i )t are linearly independent, we can equate coefficients of like terms on left and right:
−4k1= 1, ((1 + 2i )2− 4)k2= 65 2 , ((1 − 2i ) 2 − 4)k3= 65 2 . Thus, k1= −1/4 and k2= 65 2((1 + 2i )2− 4)= 65 2(1 + 4i − 4 − 4)= 65 2(4i − 7)= 1 2(−4i − 7)
and (either by a like computation or because it must be the complex con-jugate3) k3= 1 2(4i − 7). This gives p(t ) = −1 4+ 1 2(−4i − 7)e (1+2i )t+1 2(4i − 7)e (1−2i )t = −1 4− 4i 2e t (e2i t− e−2i t) −7 2e t (e2i t+ e−2i t) = −1 4+ 4e t sin 2t − 7etcos 2t .
The general solution of (N) then is the sum of H (t ) and p(t ):
x = H(t) + p(t) = c1e2t+ c2e−2t−
1 4+ 4e
t
sin 2t − 7etcos 2t .
This completes the example. We note that when using trigonometric functions as in the text, one has to solve a 2 × 2 system of equations to find k2and k3, while with complex exponentials one instead computes
the reciprocal of a complex number.
4. LAPLACE TRANSFORMS
The definition of the Laplace transform in the textbook [1, page 412] (see also [1, Example 5.2.1] and [1, Example 5.2.4]) gives
L [tn eλt] = Z ∞ 0 e−sttneλtd t = Z ∞ 0 tne−(s−λ)td t ¡integration by parts if n ≥ 1: u=tn, d v=e−(s−λ)td t ¢ = 1 s − λ if n = 0 0 + n s − λ Z ∞ 0 tn−1e−(s−λ)td t if n ≥ 1 = 1 s − λ if n = 0 n s − λL [t n−1eλt] if n ≥ 1.
3Don’t use the “must be the complex conjugate”-shortcut if you don’t know why it’s true!
Applying this recursively, we find (1) L [tneλt] = n! (s − λ)n+1 L −1 · 1 (s − λ)n ¸ = 1 (n − 1)!t n−1eλt
This will go quite far when combined with partial fractions. Note that for
n = 0 this gives the Laplace transform of an exponential function and for λ = 0 it gives the Laplace transform of a power of t.
5. PARTIAL FRACTIONS
Partial-fractions decompositions are easier with complex numbers, be-cause irreducible quadratics are no longer a difficulty: everything reduces to linear factors. One just has to patiently work the partial-fractions de-compositions that arise and put up with complex coefficients.
20. Example ([1, Example 5.3.3]: Distinct roots). Solve the initial-value problem
x0− x = 2 sin t x(0) = 0.
Note first that the method of undetermined coefficients tells us to expect no terms other than et, cos t and sin t in the solution (or: no terms other than et, ei t and e−i t). This was the purpose of Remark 19.
ApplyingL we get (s − 1)L [x] = 2Lhi 2(e −i t− ei t)i = i s + i − i s − i = 2 (s − i )(s + i ) and hence x = L−1 · 2 (s − 1)(s − i )(s + i ) ¸ .
To find x we look for the partial-fractions decomposition ofL [x]: 2 (s − 1)(s + i )(s − i )= A s − 1+ B s − i + C s + i
To determine A, B , and C , clear fractions and then insert the values of s for which the original denominator is zero, that is, s = 1,±i .
2 = A(s + i )(s − i ) + B(s − 1)(s + i ) +C (s − 1)(s − i ) For s = 1 this gives 2 = A(1 + i )(1 − i ) = A(1 − (i2)) = 2A, so A = 1.
For s = i this gives
2 = B(i − 1)(i + i ) = 2B(i − 1)i = 2B(i2− i ) = 2B(−1 − i ), so B = 1 −1 − i = − 1 2+ i 2. Finally, for s = −i we get
so C = 1 −1 + i = 1 −1 + i −1 − i −1 − i = −1 − i (−1)2− i2= − 1 2− i 2.
21. Remark. Time-saver: B and C are complex conjugates. This is no accident!
We thus obtain the partial-fractions decomposition 2 (s − 1)(s2+ 1)= 1 s − 1+ −12+i2 s − i + −12−2i s + i
Note that the two summands involving i are complex conjugates of each other, so they sum to a real number. This is why we expect their numera-tors B and C to be complex conjugates.
Now we “untransform” and sort terms:
x = L−1 " 1 s − 1+ −12+2i s − i + −12−i2 s + i # = et+¡ −1 2+ i 2¢e i t +¡ −1 2− i 2¢e −i t = et−1 2(e i t + e−i t) +i 2(e i t − e−i t) = et− cos t − sin t
We conclude that the initial-value problem
D x − x = 2sin t x(0) = 0.
has the unique solution
x = et− cos t − sin t .
The form matches with our expectations from the method of undeter-mined coefficients (Remark 19), so one only needs to check the initial condition to verify that this is correct: x(0) = e0− cos 0 − sin 0 = 1 − 1, as required.
22. Example ([1, Example 5.6.3]: Pair of double complex roots). FindL−1h s (s2+ 1)2 i . We rewrite s (s2+ 1)2= s (s − i )2(s + i )2
and produce the partial-fractions decomposition:
s (s − i )2(s + i )2= A s − i + B (s − i )2+ C s + i + D (s + i )2 becomes s = A(s − i )(s + i )2+ B(s + i )2+C (s + i )(s − i )2+ D(s − i )2.
Set s = i to get
i = B(2i )2= −4B, hence B = −i 4. Set s = −i to get
−i = D(−2i )2= −4D, hence D = i 4. Before going on we consolidate the corresponding terms:
B (s + i )2+ D(s − i )2= −i 4(s + i ) 2 +i 4(s − i ) 2 =i 4¡ − (s 2+ 2si − 1) + (s2− 2si − 1)¢ = −i 4(4si ) = s. Therefore we can rewrite
s = A(s − i )(s + i )2+ B(s + i )2+C (s + i )(s − i )2+ D(s − i )2 as
0 = A(s − i )(s + i )2+C (s + i )(s − i )2. This means that we can (and hence must!) take A = C = 0.
Therefore, L−1h s (s − i )2(s + i )2 i = L−1 h − i /4 (s − i )2+ i /4 (s + i )2 i = −i 4t e i t +i 4t e −i t= t 2 hei t− e−i t 2i i = t 2sin t . 23. Example ([1, Example 5.6.5]: Application to initial-value problem). Solve
(D2+ 1)x = cos t , x(0) = x0(0) = 0. ApplyL to both sides to get
(s2+ 1) | {z } =(s−i )(s+i ) L [x] =1 2L [e i t + e−i t] =1 2 ³ 1 s − i + 1 s + i ´ = s (s − i )(s + i ) Thus, x = L−1h s (s − i )2(s + i )2 i = t
2sin t using the previous example. 24. Example ([1, Example 5.6.4]: Same pair of double complex roots). FindL−1h 1 (s2+ 1)2 i . We write 1 (s2+ 1)2 = 1
(s − i )2(s + i )2 and produce the partial-fractions
de-composition: 1 (s − i )2(s + i )2= A s − i + B (s − i )2+ C s + i + D (s + i )2
becomes
1 = A(s − i )(s + i )2+ B(s + i )2+C (s + i )(s − i )2+ D(s − i )2.
This can be solved for A, B , C and D by multiplying out the right-hand side, sorting by powers of s and comparing coefficients to get 4 linear equations in these 4 unknowns. We try sampling useful s-values first to break the problem up.
Set s = i to get
1 = B(2i )2= −4B, hence B = −1 4. Set s = −i to get
1 = D(−2i )2= −4D, hence D = −1 4. Consolidate: B (s + i )2+ D(s − i )2= −1 4¡(s 2 + 2si − 1) + (s2− 2si − 1)¢ = −1 2(s 2 − 1). Therefore, 1 = A(s − i )(s + i )2+ B(s + i )2+C (s + i )(s − i )2+ D(s − i )2 = A(s − i )(s + i )2+C (s + i )(s − i )2−1 2(s 2− 1)
and, adding12(s2− 1) to both sides, 1 2(s 2 + 1) = 1 +1 2(s 2 − 1) = A(s − i )(s + i )2+C (s + i )(s − i )2 = A(s2+ 1)(s + i ) +C (s2+ 1)(s − i ). Dividing by the common factor s2+ 1, this becomes
1
2= A(s + i ) +C (s − i ) = (A +C )s + i (A −C )| {z }
sorted by powers of s
.
Comparing coefficients of like powers of s we find that
A +C = 0 and 1 2= i (A −C ) = 2i A, so A = 1 4i and C = − 1 4i.
Inserting these into the partial-fractions decomposition we find 1 (s − i )2(s + i )2= A s − i + B (s − i )2+ C s + i + D (s + i )2 =1/4i s − i − 1/4 (s − i )2− 1/4i s + i − 1/4 (s + i )2
This gives us L−1h 1 (s2+ 1)2 i = L−1 h1/4i s − i − 1/4 (s − i )2− 1/4i s + i − 1/4 (s + i )2 i = 1 4ie i t −t 4e i t − 1 4ie −i t−t 4e −i t =1 2 ei t− e−i t 2i − t 2 ei t+ e−i t 2 =1 2sin t − t 2cos t .
25. Example (Initial-value problem producing triple complex roots). Solve (D2+ 9)2x = −4sin3t, x(0) = x0(0) = x00(0) = 0, x000(0) = 1.
We first note that we should expect only the functions sin 3t , cos 3t ,
t sin 3t , t cos 3t , t2sin 3t , t2cos 3t in the ultimate solution (Remark 19). ApplyL to get (s2+ 9)2L [x] − 1 = −4 3 s2+ 9, so (2) L [x] = 1 (s2+ 9)2− 12 (s2+ 9)3= =(s2+9)−12 z }| { s2− 3 (s2+ 9 | {z } =(s−3i )(s+3i ) )3
We need to decompose the right-hand side as follows:
s2− 3 (s − 3i )3(s + 3i )3= A s − 3i+ B s + 3i+ C (s − 3i )2+ D (s + 3i )2+ E (s − 3i )3+ F (s + 3i )3. Clear fractions to get
s2− 3 =A(s − 3i )2(s + 3i )3+ B(s − 3i )3(s + 3i )2 +C (s − 3i )(s + 3i )3+ D(s − 3i )3(s + 3i ) + E(s + 3i )3+ F (s − 3i )2.
For s = 3i this becomes
E =−9 − 3 (6i )3 = 2 · 6 6 · 3 · 2 · 6i = 1 18i, and F must be the complex conjugate: F = − 1
18i. We consolidate these 2 terms first: E (s − 3i )3+ F (s + 3i )3 = 1/18i (s − 3i )3− 1/18i (s + 3i )3=
=(s+3i )318i −(s−3i )318i
z }| {
s2− 3 (s2+ 9)3.
Note from equation (2) above that we have just stumbled upon the partial-fractions decomposition we need, that is, that (using equation (1)) we have x = L−1 · s2− 3 (s2+ 9)3 ¸ = 1 18iL −1 · 1 (s − 3i )3− 1 (s + 3i )3 ¸ = 1 18 t2e3i t− t2e−3i t 2i .
In short, the solution to the given initial-value problem is
x = 1
18t
2
sin 3t .
By inspection (really!) one can see that this satisfies the initial condition. This last example illustrates that we can in this way handle complex roots of any multiplicity, not just double complex roots.
REFERENCES
[1] Martin M. Guterman, Zbigniew H. Nitecki, Differential Equations – A First Course, 3rded., Saunders (1992).
