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C HAPTER 5: L INEAR D IFFERENTIAL
E QUATIONS
S ECTION 5.3 — C OMPLEX S OLUTIONS
Dr. Bisher M. Iqelan
[email protected]
Department of Mathematics
The Islamic University of Gaza
2016-2017, Semester 2
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Dr. Bisher M. Iqelan
Lecture 25: Sec 5.3 — Complex Solutions
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B ASIC P ROPERTIES OF C OMPLEX NUMBERS
FACT
1 A complex number z = a + ib, where a is a real part and b is
√
an imaginary part and i = −1.
2
Any real number a is complex since a = a + 0i, so R ⊆ C.
3
z1 = a + ib, z2 = c + id are equal iff a = c, and b = d.
4
A complex function w (x ) = u (x ) + iv (x ) and its derivative
w 0 (x ) = u 0 (x ) + iv 0 (x ), for example if:
w (x ) = e2x + i ln(2x ), then w 0 (x ) = 2e2x + i x1 .
5
If w (x ) = u (x ) + iv (x ) is a complex solution of the
homogeneous DE L[y ] = 0, then both u (x ) and v (x ) are real
solutions of the same DE L[y ] = 0.
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Lecture 25: Sec 5.3 — Complex Solutions
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E ULER ’ S F ORMULA
FACT (E ULER ’ S FORMULA : S OME PROPERTIES )
1
Euler’s formula: The exponential function ei θ is written as
ei θ = cos θ + i sin θ .
2
e−i θ = ei (−θ ) = cos(−θ ) + i sin(−θ ) ⇒ e−i θ = cos θ − i sin θ.
3
ex +iy = ex .eiy = ex (cos y + i sin y ) = ex cos y + iex sin y
4
ei π = cos π + i sin π = −1.
5
For any real number b,
eibx + e−ibx = cos bx + i sin bx + cos bx − i sin bx
i
1 h ibx
=2 cos bx ⇒ cos bx =
e + e−ibx
2
Similarly, sin bx = 2i1 eibx − e−ibx .
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Lecture 25: Sec 5.3 — Complex Solutions
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n − TH O RDER H OMOGENEOUS DE
For the (Hc) DE,
0
a0 y (n) + a1 y (n−1) + · · · + an−1 y + an y = 0
(Hc)
The auxiliary (characteristic) polynomial of the DE (Hc) is
a0 r n + a1 r n−1 + · · · + an−1 r + an = 0
(ChP)
If the roots of (Hc) are complex, then every pair are conjugate,
that is, if r1 = α + i β then r2 = α − i β.
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Lecture 25: Sec 5.3 — Complex Solutions
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p (D ) H AS T WO C ONJUGATE C OMPLEX R OOTS α ± i β
Suppose p (D ) has two conjugate complex roots
r1 = α + i β,
r2 = α − i β,
α, β ∈ R
If we slightly extend our discussion to complex-valued DE, it is not
hard to see that the previous method works again and we get a
general (complex-valued) solution
y = c1 er1 x + c2 er2 x
c1 , c2 ∈ C
Still we need to get back to the real domain ... So, lets do some
further manipulation by using the fact that
ei θ = cos θ + i sin θ
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Lecture 25: Sec 5.3 — Complex Solutions
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p (D ) H AS T WO C ONJUGATE C OMPLEX R OOTS α ± i β
The general solution to Ly = 0 where L = p (D ) is
y =C1 e(α+i β)x + C2 e(α−i β)x = C1 eαx ei βx + C2 eαx e−i βx
=C1 eαx (cos βx + i sin βx ) + C2 eαx (cos βx − i sin βx )
=(C1 + C2 )eαx cos βx + i (C1 − C2 )eαx sin βx
To get a real-valued solution, there are two choices:
• Pick C1 + C2 = 1, C1 − C2 = 0 : we get
y = f1 (x ) = eαx cos βx
• Pick C1 + C2 = 0, C1 − C2 = −i : we get
y = f2 (x ) = eαx sin βx
Since f1 and f2 are linearly independent, the general real-valued
solution to Ly = 0 where L = p (D ) is
c1 eαx cos βx + c2 eαx sin βx ,
c1 , c2 ∈ R.
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Lecture 25: Sec 5.3 — Complex Solutions
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E XAMPLE
Find the general solution of the DE
(D 2 + 4D + 5)y = 0
Answer: Let y = erx be a solution, then
Dy = r erx ,
D 2 y = r 2 erx
The characteristic equation is r 2 + 4r + 5 = 0, solving we have
√
−4 ± 2i
−4 ± 16 − 4.1.5
=
= −2 ± i
r=
2
2
The general solution is
y = e−2x [c1 cos x + c2 sin x ]
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Lecture 25: Sec 5.3 — Complex Solutions
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E XAMPLE
Find the general solution of the DE
(D 5 − 16D )y = 0
Answer: Let y = erx be a solution, then the characteristic
equation is r 5 − 16r = 0, solving we have
r 5 − 16r = r (r 4 − 16) = 0
⇒ r (r 2 − 4)(r 2 + 4) = r (r − 2)(r + 2)(r 2 + 4)0
⇒ r = 0, r = 2, r = −2, r = ±2i
The general solution is
y = c1 + c2 e2x + c3 e−2x + c4 cos 2x + c5 sin 2x
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Lecture 25: Sec 5.3 — Complex Solutions
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E XAMPLE
Solve the IVP of the DE
y 00 + 16y = 0, y (0) = 2, y 0 (0) = −2
Answer: Let y = erx be a solution, then the characteristic
equation is r 2 + 16 = 0, solving we have r = ±4i
The general solution is
y =c1 cos 4x + c2 sin 4x,
y (0) = 2 ⇒ c1 = 2
y 0 = − 4c1 sin 4x + 4c2 cos 4x,
y 0 ( 0 ) = − 2 ⇒ c2 = −
1
2
Thus the particular solution is
y = 2 cos 4x −
1
sin 4x
2
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Lecture 25: Sec 5.3 — Complex Solutions
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H OME W ORK S EC .5.3
HW. (E X . S EC .5.3)
Solve:
Q3 (a, b, c ), Q7 , Q8 .
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