Download General solutions, independence of functions and

Today
• General solutions, independence of functions and the Wronskian
• Distinct roots of the characteristic equation
• Review of complex numbers
• Complex roots of the characteristic equation
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Saturday, 17 January, 15
Homog. eq. with constant coeff. (Section 3.1)
• Which of the following functions are also solutions?
(A) y(t) = y1(t)2
(B) y(t) = y1(t)+y2(t)
(C) y(t) = y1(t) y2(t)
(D) y(t) = y1(t) / y2(t)
Last class, we found that if y1(t)
is a solution to
00
0
ay + by + cy = 0
then so is y(t) = C1y1(t).
• In fact, the following are all solutions: C1y1(t), C2y2(t), C1y1(t)+C2y2(t).
• With first order equations, the arbitrary constant appeared through an
integration step in our methods. With second order equations, not so lucky.
• Instead, find two independent solutions, y1(t), y2(t), by whatever method.
• The general solution will be y(t) = C1y1(t) + C2y2(t).
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Saturday, 17 January, 15
Homog. eq. with constant coeff. (Section 3.1)
• One case where the arbitrary constants DO appear as we calculate:
y 00 + y 0 = 0
y + y = C1
0
et y 0 + et y = C1 et
(et y)0 = C1 et
et y = C1 et + C2
y = C1 + C2 e
t
• More common would be that we find solutions y(t) = 1 and y(t)= e-t and simply
write down
t
y = C1 + C2 e
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Saturday, 17 January, 15
Homog. eq. with constant coeff. (Section 3.1)
• So in general how do we find the two independent solutions y1 and y2?
• Exponential solutions seem to be common so let’s assume y(t)=ert and see if
that gets us anything useful..
• Solve
y 00 + y 0 = 0
by assuming y(t) = ert for some constant r.
(e ) + (e ) = 0
rt 00
rt 0
r2 ert + rert = 0
2
r +r =0
r(r + 1) = 0
r = 0, r = 1
y = C1 e + C2 e
0
y = C1 + C2 e
t
t
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Saturday, 17 January, 15
Homog. eq. with constant coeff. (Section 3.1)
• Solve
y
00
4y = 0
subject to the ICs
(A) y(t) = C1 e
2t
(B) y(t) = 2e
y(0) = 3, y (0) = 2
0
+ C2 e
.
2t
+e
7 4t 5 4t
(C) y(t) = e + e
4
4
2t
2t
y(t)
=
e
+
2e
(D)
2t
(E) y(t) = C1 e
4t
2t
+ C2 e
4t
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Saturday, 17 January, 15
Homog. eq. with constant coeff. (Section 3.1)
• For the general case,
ay + by + cy = 0, by assuming y(t) = e
00
0
rt
we get the characteristic equation:
ar + br + c = 0
2
• There are three cases.
i. Two distinct real roots: b2 - 4ac > 0. ( r1 ≠ r2 )
ii.A repeated real root: b2 - 4ac = 0.
iii.Two complex roots: b2 - 4ac < 0.
r1 t
y
(t)
=
e
• For case i, we get 1
and
y2 (t) = er2 t.
• Do our two solutions cover all possible ICs? That is, can we use them to
form a general solution?
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Saturday, 17 January, 15
Independence and the Wronskian (Section 3.2)
2t
2t+3
y
(t)
=
e
y
(t)
=
e
• Example: Suppose 1
and 2
3
are two
solutions to some equation. Can we solve ANY initial
condition .y(0) = y0 , y 0 (0)
. = v0 with these two solutions?
y(t) = C1 e2t+3 + C2 e2t
y(0) = C1 e3 + C2 e
y 0 (0) = 2C1 e3 + 2C2 e
3
3
= y0
3
= v0
• Solve this system for C1, C2...
• Can’t do it. Why?
Saturday, 17 January, 15
✓
◆✓ ◆ ✓ ◆
e
e
C1
y0
=
3
3
2e 2e
C2
v0
✓ 3
◆
e
e 3
det
=0
3
3
2e 2e
3
3
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Independence and the Wronskian (Section 3.2)
• For any two solutions to some linear ODE, to ensure that we have a
general solution, we need to check that
✓
◆
y1 (0) y2 (0)
0
det 0
= y1 (0)y2 (0)
0
y1 (0) y2 (0)
0
y1 (0)y2 (0)
6= 0
• For ICs other than t0=0, we require that
0
y1 (t0 )y2 (t0 )
0
y1 (t0 )y2 (t0 )
6= 0
• This quantity is called the Wronskian.
W (y1 , y2 )(t) =
0
y1 (t)y2 (t)
0
y1 (t)y2 (t)
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Saturday, 17 January, 15
Independence and the Wronskian (Section 3.2)
• Two functions y1(t) and y2(t) are linearly independent provided that the
only way that C1y1(t) + C2y2(t) = 0 for all values of t is when C1=C2=0.
e.g. y1 (t) = e2t+3 and y2 (t) = e2t
3
are not independent.
Find values of C1≠0 and C2≠0 so that C1y1(t) + C2y2(t) = 0.
(A)
C1 = e
2t 3
, C2 =
(B)
C1 = e
2t+3
, C2 =
(C)
C1 = e
3
, C2 = e3
(D)
C1 = e
3
, C2 =
(E)
C1 = e3 , C2 =
e
2t+3
e
2t 3
e3
e
3
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Saturday, 17 January, 15
Independence and the Wronskian (Section 3.2)
• Two functions y1(t) and y2(t) are linearly independent provided that the
only way that C1y1(t) + C2y2(t) = 0 for all values of t is when C1=C2=0.
e.g. y1 (t) = e2t+3 and y2 (t) = e2t
3
are not independent.
• The Wronskian is defined for any two functions, even if they aren’t
solutions to an ODE.
W (y1 , y2 )(t) =
0
y1 (t)y2 (t)
0
y1 (t)y2 (t)
• If the Wronskian is nonzero for some t, the functions are linearly
independent.
• If y1(t) and y2(t) are solutions to an ODE and the Wronskian is nonzero
then they are independent and
y(t) = C1 y1 (t) + C2 y2 (t)
is the general solution. We call y1(t) and y2(t) a fundamental set of solutions
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and we can use them to solve any IC.
Saturday, 17 January, 15
Independence and the Wronskian (Section 3.2)
• So for case i (distinct roots), can we form a general solution from
y1 (t) = er1 t
y2 (t) = er2 t ?
and
• Must check the Wronskian:
W (e
r1 t
r2 t
,e
)(t) = e
r1 t
= (r1
So yes!
y(t) = C1 e
r1 t
r2 t
r1 t r 2 t
r2 e
r1 e
r2 )e
r1 t r2 t
+ C2 e
r2 t
e
e
6= 0
is the general solution.
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Saturday, 17 January, 15
Independence and the Wronskian (Section 3.2)
y 00 + 9y = 0. Find the roots of the
• Example: Consider the equation
characteristic equation (i.e. the r values).
(A) r1 = 3, r2 = -3.
(B) r1 = 3 (repeated root).
(C) r1 = 3i, r2 = -3i.
(D) r1 = 9, (repeated root).
As we’ll see soon, this means that
y1(t) = cos(3t) and y2(t)=sin(3t).
Do these form a fundamental set of
solutions? Calculate the Wronskian.
W (cos(3t), sin(3t))(t) =
(A) 0
(C) 3
(B) 1
(D) 2 cos(3t) sin(3t)
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Saturday, 17 January, 15
Distinct roots - asymptotic behaviour (Section 3.1)
• Three cases:
(i) Both r values positive.
e.g.
Except for the zero solution
y(t)=0, the limit lim y(t) ...
y(t) = C1 e2t + C2 e5t
t!1
(A) ...is unbounded for all ICs.
(ii) Both r values negative.
e.g.
y(t) = C1 e
2t
+ C2 e
5t
(iii) The r values have opposite sign.
e.g.
y(t) = C1 e
2t
+ C2 e
5t
(B) ...is unbounded for most
ICs but not for a few
carefully chosen ones.
(C) ...goes to zero for all ICs.
Challenge: come up with an initial condition
for (iii) that has a bounded solution.
Saturday, 17 January, 15
13
Distinct roots - asymptotic behaviour (Section 3.1)
• Three cases:
(i) Both r values positive.
e.g.
Except for the zero solution
y(t)=0, the limit lim y(t)...
y(t) = C1 e2t + C2 e5t
t! 1
(A) ...is unbounded for all ICs.
(ii) Both r values negative.
e.g.
y(t) = C1 e
2t
+ C2 e
5t
(iii) The r values have opposite sign.
e.g.
y(t) = C1 e
2t
+ C2 e
5t
(B) ...is unbounded for most
ICs but not for a few
carefully chosen ones.
(C) ...goes to zero for all ICs.
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Saturday, 17 January, 15
Complex roots (Section 3.3)
• Complex number review (Euler’s formula)
• Complex roots of the characteristic equation
• From complex solutions to real solutions
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Saturday, 17 January, 15
Complex number review
• We define a new number:
i=
p
1
• Before, we would get stuck solving any equation that required squarerooting a negative number. No longer.
• e.g. The solutions to
x2
4x + 5 = 0
are
x=2+i
and
x=2
i
2
ax
+ bx + c = 0 , when b2 - 4ac < 0, the solutions
• For any equation,
have the form x = ± ⇥i where α and β are both real numbers.
• For α+βi, we call α the real part and β the imaginary part.
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Saturday, 17 January, 15
Complex number review
• Adding two complex numbers:
(a + bi) + (c + di) = a + c + (b + d)i
• Multiplying two complex numbers:
(a + bi)(c + di) = ac
bd + (ad + bc)i
• Dividing by a complex number:
1
(a + bi)/(c + di) = (a + bi)
(c + di)
• What is the inverse of c+di?
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Saturday, 17 January, 15
Complex number review
• What is the inverse of c+di written in the usual complex form p+qi?
(A)
c
di
(B)
c + di
c2 + d2
(C)
(D)
c di
c2 + d2
1
c di
c di
c +d
(cd
(c + di) 2
=
2
c +d
c2 + d2
2
2
dc)i
=1
• Dividing by a complex number:
c di
ac + bd (bc ad)i
=
(a + bi)/(c + di) = (a + bi) 2
+
c + d2
c2 + d2
c2 + d2
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Saturday, 17 January, 15
Complex number review
• Definitions:
• Conjugate - the conjugate of a + bi is
a + bi = a
bi
a + bi is
p
|a + bi| = a2 + b2
• Magnitude - the magnitude of
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Saturday, 17 January, 15
Complex number review
• Geometric interpretation of complex numbers
• e.g.
a + bi
a = M cos
b = M sin
p
M = a2 + b2
✓ ◆
b
= arctan
a
a + bi = M (cos + i sin )
is sometimes called the
argument or phase of a + bi.
✓
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Saturday, 17 January, 15
Complex number review
• Toward Euler’s formula
• Taylor series - recall that a function can be represented as
00
f
(x0 )
0
2
f (x) = f (x0 ) + f (x0 )(x x0 ) +
(x x0 ) + · · ·
2!
x23 x2x45 x3
1 + x ++ + · · +
· ···
• What function has Taylor series x
3! 2!4!
5! 3!
2!
(A) cos x
(C) ex
(B) sin x
(D) ln x
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Saturday, 17 January, 15
Complex number review
• Use Taylor series to rewrite
cos + i sin
=1
cos + i sin
2
4
✓
✓
+
2!
4!
= 1 + i + ( 1)
1=i
2
=1+i +i
2
2
.
· · · +i
2
2!
✓
3
3!
+ ( 1)i
+i
3
3
+i
3
+ ( 1)
4
···
5!
2
3!
4
+
5
4
4!
◆
+ ···
+ ···
2!
3!
4!
2
3
4
(i )
(i )
(i )
i
=1+i +
+
+
+ ··· = e
2!
3!
4!
sin x = x
Saturday, 17 January, 15
x3
x5
+
3!
5!
···
cos x = 1
2
4
x
x
+
2!
4!
···
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Complex number review
• Use Taylor series to rewrite
cos + i sin
.
cos + i sin
Euler’s formula:
=e
i
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Saturday, 17 January, 15
Complex number review
• Geometric interpretation of complex numbers
• e.g.
a + bi
a = M cos
b = M sin
p
M = a2 + b2
✓ ◆
b
= arctan
a
a + bi = M (cos + i sin )
a + bi = M e
i
(Polar form makes multiplication
much cleaner)
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Saturday, 17 January, 15