Math 220: Lecture 15 § 4.3 - University of Illinois at Chicago

Math 220: Lecture 15
§ 4.3
Professor Nicholls
Department of Mathematics, Statistics,
and Computer Science
University of Illinois at Chicago
Math 220: Lecture 15
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Chapter 4: Linear Second Order Equations
Due to the importance of Newton’s Second Law (relating positions,
velocities, and accelerations) the most common (and important)
Ordinary Differential Equations (ODEs) are second order.
In contrast to the first order case, there are very few general
purpose solution techniques for second (and higher) order ODEs.
In fact, the only general case that we can solve is Linear and
Constant Coefficient!
As we shall see, our method of solution reduces the second order
ODE to a polynomial of order two.
We recall that the Quadratic Formula tells us that we get three
“types” of solutions: Distinct real, repeated real, and imaginary.
For this reason, our solution technique boils down to these three
cases:
Distinct Real Roots [§ 4.2]
Repeated Real Root [§ 4.2]
Imaginary Roots [§ 4.3, today]
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4.3: Auxiliary Equations with Complex Roots
For the moment we restrict our attention to second order, linear,
constant coefficient homogeneous ODEs. The most general such
equation can be written:
ay 00 + by 0 + cy = 0,
a, b, c ∈ R.
Recall, our “method” is an inspired guess
y (t) = ert ,
where r is a constant to be determined.
Inserting our guess into the ODE gives:
ar 2 ert + brert + cert = 0,
while factoring yields:
p(r )ert := ar 2 + br + c ert = 0.
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The Method
Either the polynomial p(r ) = ar 2 + br + c or the exponential is
zero. The latter cannot occur so we must have the former:
p(r ) = 0.
So, the key is to find roots (zeros) of the “characteristic equation
(polynomial)”:
p(r ) = ar 2 + br + c.
In the quadratic case we have the Quadratic Formula:
p
r = −b ± b2 − 4ac /(2a).
There are three “cases” decided by the discriminant
∆ = b2 − 4ac.
∆ > 0: Distinct real roots.
∆ = 0: Repeated real root.
∆ < 0: Imaginary (complex conjugate) roots.
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The Method: Complex Roots
The method worked immediately when the roots were real and
distinct (r1 6= r2 ∈ R):
y (t) = C1 er1 t + C2 er2 t ,
and we found solutions for the real, repeated case
(r1 = r2 = r ∈ R):
y (t) = C1 ert + C2 tert .
In the complex case we have conjugate roots:
p
r = α ± iβ, α = −b/(2a), β = 4ac − b2 /(2a),
but what does e(α+iβ)t mean?
We can get a start by assuming that the rules of exponentials of
sums still hold for complex numbers
e(α+iβ)t = eαt eiβt ,
and the first term makes sense.
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Example # 1
Example #1: Find solutions of
y 00 + y = 0.
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Work Space
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Solution: Example # 1
Solution #1: Making the guess we advocated earlier:
y (t) = ert ,
and inserting this into the ODE gives:
r 2 ert + ert = 0.
Factoring we find:
r 2 + 1 ert = 0,
and the characteristic polynomial is
p(r ) = r 2 + 1.
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Solution: Example # 1, cont.
We have found the characteristic polynomial to be
p(r ) = r 2 + 1,
and we can find the roots from the Quadratic Formula (QF):
q
r = (0 ± 02 − 4(1)(1))/(2(1)) = ±i,
so that r1 = i and r2 = −i.
Question: The ODE can be rewritten as y 00 = −y . What function(s)
are equal to the opposite of their second derivatives?
Answer: Sines and cosines!
Solution: Pick
y1 (t) = cos(t), y2 (t) = sin(t).
Independent?
W = y1 y20 − y10 y2 = cos(t) cos(t) − (− sin(t)) sin(t) = 1.
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Euler’s Formula
Remark 1: So, the complex exponential eit must be related to the
trigonometric functions: sine and cosine!
The “highlight” of Math 220: Recall the Taylor expansion for the
exponential:
x2 x3 x4 x5
+
+
+
+ ....
2!
3!
4!
5!
Follow Euler and simply evaluate at x = (it):
ex = 1 + x +
(it)2 (it)3 (it)4 (it)5
+
+
+
+ ....
2!
3!
4!
5!
Now, gather things up into real and imaginary parts:
t2
t4
t3
t5
it
e = 1−
+
+ ... + i t −
+
+ ... .
2! 4!
3! 5!
eit = 1 + it +
But we know these Taylor series: Sines and cosines!
eit = cos(t) + i sin(t).
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Euler’s Formula
We now have Euler’s Formula:
eit = cos(t) + i sin(t).
We can generalize our arguments (for application to the problems we
wish to solve):
e(α+iβ)t = eαt (cos(βt) + i sin(βt)) = eαt cos(βt) + ieαt sin(βt).
Notice that we can use the even/odd properties of cosine/sine to
deduce:
e(α−iβ)t = eαt (cos(βt) − i sin(βt)) = eαt cos(βt) − ieαt sin(βt).
Remark 2: My favorite formula in mathematics can be derived by
setting t = π in Euler’s formula (and rearranging):
eiπ + 1 = 0,
which relates the five most important numbers in mathematics!
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The Method: Complex Roots, cont.
In the complex case we have r = α ± iβ, and from these we can
construct two solutions:
y1 (t) = eαt cos(βt),
y2 (t) = eαt sin(βt).
These solutions are independent:
W = eαt cos(βt) αeαt sin(βt) + βeαt cos(βt)
− αeαt cos(βt) − βeαt sin(βt) eαt sin(βt)
= e2αt β cos2 (βt) + sin2 (βt) = βe2αt .
The General Solution in this case is:
y (t) = C1 eαt cos(βt) + C2 eαt sin(βt).
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Example # 2
Example #2: (NSS 4.3 # 6): Find the general solution of
w 00 + 4w 0 + 6w = 0.
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Work Space
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Solution: Example # 2
Solution #2: Making the guess we advocated earlier:
w(t) = ert ,
and inserting this into the ODE gives:
r 2 ert + 4rert + 6ert = 0.
Factoring we find:
r 2 + 4r + 6 ert = 0,
and the characteristic polynomial is
p(r ) = r 2 + 4r + 6.
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Solution: Example # 2, cont.
We have found the characteristic polynomial to be
p(r ) = r 2 + 4r + 6,
and we can find the roots from the Quadratic Formula (QF):
q
√
2
r = −4 ± (4) − 4(1)(6) /(2(1)) = −2 ± i 2,
√
√
so that r1 = −2 + i 2 and r2 = −2 − i 2.
Thus, we have two solutions
√
√
w1 (t) = e−2t cos( 2t), w2 (t) = e−2t sin( 2t),
and propose the “General Solution”:
w(t) = C1 w1 (t) + C2 w2 (t)
√
√
= C1 e−2t cos( 2t) + C2 e−2t sin( 2t).
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Example # 3
Example #3: (NSS 4.3 # 22): Solve the initial value problem:
y 00 + 2y 0 + 17y = 0,
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y (0) = 1,
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y 0 (0) = −1.
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Work Space
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Solution: Example # 3
Solution #3: Making the guess we advocated earlier:
y (t) = ert ,
and inserting this into the ODE gives:
r 2 ert + 2rert + 17ert = 0.
Factoring we find:
r 2 + 2r + 17 ert = 0,
and the characteristic polynomial is
p(r ) = r 2 + 2r + 17.
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Solution: Example # 3, cont.
We have found the characteristic polynomial to be
p(r ) = r 2 + 2r + 17,
and we can find the roots from the Quadratic Formula (QF):
q
r = −2 ± (2)2 − 4(1)(17) /(2(1)) = −1 ± i4,
so that r1 = −1 + i4 and r2 = −1 − i4.
Thus, we have two solutions
y1 (t) = e−t cos(4t),
y2 (t) = e−t sin(4t),
and propose the “General Solution”:
y (t) = C1 y1 (t) + C2 y2 (t)
= C1 e−t cos(4t) + C2 e−t sin(4t).
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Example # 3, cont.
We finish by using the initial conditions:
1 = y (0) = C1 e0 cos(0) + C2 e0 sin(0) = C1 ,
and
− 1 = y 0 (0) = C1 −e0 cos(0) − 4e0 sin(0)
+ C2 −e0 sin(0) + 4e0 cos(0) = −C1 + 4C2 ,
so that
C2 = (−1 + 1)/4 = 0.
So, we have the unique solution
y (t) = e−t cos(4t).
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Example # 4
Example #4: (NSS 4.3 # 15): Find the general solution of
y 00 + 10y 0 + 41y = 0.
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Work Space
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Solution: Example # 4
Solution #4: Making the guess we advocated earlier:
y (t) = ert ,
and inserting this into the ODE gives:
r 2 ert + 10rert + 41ert = 0.
Factoring we find:
r 2 + 10r + 41 ert = 0,
and the characteristic polynomial is
p(r ) = r 2 + 10r + 41.
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Solution: Example # 4, cont.
We have found the characteristic polynomial to be
p(r ) = r 2 + 10r + 41,
and we can find the roots from the Quadratic Formula (QF):
q
r = −10 ± (10)2 − 4(1)(41) /(2(1)) = −5 ± i4,
so that r1 = −5 + i4 and r2 = −5 − i4.
Thus, we have two solutions
y1 (t) = e−5t cos(4t),
y2 (t) = e−5t sin(4t),
and propose the “General Solution”:
y (t) = C1 y1 (t) + C2 y2 (t)
= C1 e−5t cos(4t) + C2 e−5t sin(4t).
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Example # 5
Example #5: (NSS 4.3 # 24): Solve the initial value problem:
y 00 + 9y = 0,
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y (0) = 1,
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y 0 (0) = 1.
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Work Space
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Solution: Example # 5
Solution #5: Making the guess we advocated earlier:
y (t) = ert ,
and inserting this into the ODE gives:
r 2 ert + 9ert = 0.
Factoring we find:
r 2 + 9 ert = 0,
and the characteristic polynomial is
p(r ) = r 2 + 9.
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Solution: Example # 5, cont.
We have found the characteristic polynomial to be
p(r ) = r 2 + 9,
and we can find the roots from the Quadratic Formula (QF):
q
2
r = 0 ± (0) − 4(1)(9) /(2(1)) = ±i3,
so that r1 = i3 and r2 = −i3.
Thus, we have two solutions
y1 (t) = cos(3t),
y2 (t) = sin(3t),
and propose the “General Solution”:
y (t) = C1 y1 (t) + C2 y2 (t)
= C1 cos(3t) + C2 sin(3t).
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Solution: Example # 5, cont.
We finish by using the initial conditions:
1 = y (0) = C1 cos(0) + C2 sin(0) = C1 ,
and
1 = y 0 (0) = C1 (−3 sin(0)) + C2 3 cos(0) = 3C2 ,
so that
C2 = 1/3.
So, we have the unique solution
y (t) = cos(3t) + (1/3) sin(3t).
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