1
Lecture 17
Finally, we remark that, as in the case of a real r, for complex r;
D [erx ] = rerx :
Example 1 Let r = a + bi. Then
d ax ibx
d (a+bi)x
=
e
ee
dx
dx
d ax
=
[e (cos bx + i sin bx)]
dx
= aeax (cos bx + i sin bx) + eax ( b sin bx + bi cos bx)
=
1
Lecture 29
Example 1 (Continue) f (t) = jsin tj ;
t
: It is evident that f (t)
is an even function. Hence, bn = 0; n = 1; 2; : , and For n > 1,
Z
2
sin t cos ntdt =
an =
0
=
1
2
n1
2
n+1
0
Recall that
2
a0 =
Z
=
4
if
(n+1)(n 1)
n is even
0 = 0 if n is o
1
Lecture 28
Z
Z
1
sin mt sin ntdt =
2
Z
1
=
2
[cos (m
n) t + cos (m + n) t] dt
Z
1
n) tdt +
cos (m + n) tdt = 0:
2
cos (m
Let m; n = 1; 2; : . Then
Z
Z
1
sin mt cos ntdt =
2
Z
1
=
2
[sin (m + n) t + sin (m n) t] dt
Z
1
sin (m n) tdt +
sin (m + n) tdt = 0
1
Lecture 27
1.1
Periodic functions
A function y = f (t), 1 < t < +1, is said to be periodic if there is a
positive number p such that
f (t) = f (t + p) ; for every t.
Such smallest positive number p, when exists, is called the period of f . Functions sin
1
1.1
Lecture 26
The whirling string
We begin by recalling basic facts about circular motion. 1.1.1 Circular motion
Suppose that a point is moving along the circle x2 + y 2 = R2 . The position vector ! (t) is given by r ! (t) = (R cos !t; R sin !t) r (the
1
Lecture 24
1.1
Damped forced oscillation
Recall the equation of the damped forced oscillation:
mx00 + cx0 + kx = F (t) :
We assume that F (t) = F0 cos !t. Recall also that for complimentary solution
mx00 + cx0 + kx = 0
the following cases are possible:
1
1.1
Lecture 23
Undamped forced oscillation
General equation:
mx00 + kx = F0 cos !t:
Recall that
r
k
m
is called the natural frequency of the system. In what follows, we assume
that ! 6= ! 0 . The case ! = ! 0 is called resonance and will be considered
l
1
1.1
Let
Lecture 22
Variation of parameters
L [y] = y 00 + py 0 + qy;
where p and q are constants. Consider nonhomogeneous equation: L [y] = f and complimentary equation L [y] = 0: Let y1 and y2 be a fundamental set of solutions of the complimentary equ
1
Lecture 21
L [y1p ] = 2 cos 3x; L [y2p ] = 3 sin 3x:
Example 1 (Continue)
Then y1p (x) = x (A1 cos 3x + B1 sin 3x) and y2p (x) = x (A2 cos 3x + B2 sin 3x) ) yp (x) = y1p (x) + y2p (x) = x (A1 cos 3x + B1 sin 3x) + x (A2 cos 3x + B2 sin 3x) = x [(A1 + A2
1
1.1
Lecture 20
Nonhomogeneous equations with constant coe cients
We consider the dierential operator
L = D n + p1 D n
1
+ p2 D n
2
+ : + pn 1 D + pn D0 ;
where pj = const; j = 1; 2; 3; :; n.
Recall that a nonhomogeneous ODE with constants coe cients i
1
Lecture 30
Example 1 (Continue)
f (t) =
0 if
5t0
:
1 if 0 < t < 5
We have:
a0 =
1
5
Z
L = 5; p = 10:
Z
Z
10
15
f (t) dt =
0dt +
dt = 1:
55
50
5
5
Example 2 (Continue)
Z
Z
15
nt
15
nt
sin n
an =
f (t) cos
dt =
cos
dt =
= 0; n = 1; 2; : .
55
5
50
5
n
Z
Z
3

130
36
kdx = kx + c
x dx ( 1) =
( k
)
1 x +1 + c +1
1 1 dx =  + c 2 x x
,
x dx =
2 2 x +c 3
3
,
1 x
dx = 2 x + c
1 dx = ln  x  + c x
ax +c ln a
, e x dx = e x + c
a x dx =
sin x dx =  cos x + c
cos x dx = sin x + c
sin
1
2
x
dx =
csc x d
1
Lecture 16
Recall that we are looking for solution in the form y = u (x) er0 x . We showed that u(m0 ) er0 x = 0, whence u(m0 ) (x) 0 ) u (x) = C0 + C1 x + : + Cm0 1 xm0 1 :
1
So, if r = r0 is a root of multiplicity m0 , then y (x) = C0 + C1 x + : + Cm0
1
Lecture 15
Example 1 (Continue) Consider the following initial value problem:
y 000
6y 00 + 11y 0
6y = 0;
y (0) = 0; y 0 (0) = 0 and y 00 (0) = 3:
General solution is given by
y (x) = C1 ex + C2 e2x + C3 e3x :
Now we employ the initial conditions: y (0)
1
Lecture 14
Example 1 f (x) = cos x; g (x) = sin x; 0 x . Here, intuitively, f (x) and g (x) should be independent. To justify our guess, consider f (x) + g (x) 0 ) cos x + sin x = 0; 0 x In particular, for x = 0; + So, we have sin x = 0; 0 Take x = =2.
1
Lecture 13
If the functions f (x) and g (x) ; a < x < b; are linearly dependent on the interval a < x < b, then their Wronskian vanishes, for every x; a < x < b : W (f; g) jx = 0; a < x < b: If Wronskian of functions f and g is not zero at least at one
1
1.1
Lecture 11
Escape velocity
What initial velocity is needed to escape the Earth? Here we give the answer
on that question. We have
dv
=
dt
GME
,
y2
where y is the distance from the body to the center of Earth.
By the Chain Rule,
dv
dv dy
dv
=
=v :
dt
1
Lecture 10
Example 1 (Continue) Consider a population P (t) of unsophisticated animals that rely solely on chance of encounter to meet mates for reproductive purposes. We can assume that such encounter occurs at a rate that is proportional to the produc
1
Lecture 9
Now we derive the solution of the following initial value problem for logistic
equation:
P 0 = kP (M P )
Z
Z
dP
= k dt:
P (M P )
1
1
=
P (M P )
M
So,
Z
1
M
P
+
1
P
:
Z
Z
1
dP
1
dP
1
dP
=
+
=
[ln P
P (M P )
M
MP
M
P
M
1
P
ln
:
M
MP
ln jM
Finall
1
Lecture 8
Theorem 1 (Criterion for exactness) Suppose that the functions M (x; y) and N (x; y) are continuous and have continuous .rst order partial derivatives in the open rectangle R: a < x < b; c < y < d. Then the dierential equation M dx + N dy = 0
1
1.1
Lecture 7
Homogeneous equations
y ; x 6= 0: x
Equation y 0 = f (x; y) is called homogeneous if f (x; y) = F
In other words, the rate function of the homogeneous equation depends on y=x rather then on x and y. To solve a homogeneous equation, conside
Exam 3: Math. 285 (C1)
April 23, 2012
Version: E
You cannot use calculator during the exam.
Please read all questions carefully, answer all parts of all questions.
Solutions without arguments and intermediate calculations will not be accepted.
Partial cre
1
Lecture 19
Example 1 (Continue) Two pendulums are of length L1 and L2 , and located at the respected distances R1 and R2 from the center of Earth, have periods p1 and p2 . Prove that p R 1 L1 p1 p : = p2 R 2 L2 Solution.
00
g + = 0 ) !0 = L g=G
MEarth ;
1
Lecture 21
1.1
Mechanical vibrations
A mass m is attached both to a spring (that exerts the force FS = k x) and
to a dashpot (shock absorber) that exerts a force FR = cv , where v = x0 .
Newton Second Law,
s
mx00 =
kx
cx0 )
mx00 + cx0 + kx = 0.
If we ha
1
1.1
Lecture 6
Linear ODE
We consider the rst order linear ODE:
y 0 + p (x) y = q (x) ;
where p (x) and q (x) are continuous functions, for a < x < b.
Integrating factor for a dierential equation is a function (x) such that
multiplication of each side of
Math. 285(C1): Exam 1
Monday, February 13, 2012
10am10:50am, room 245AH
Topics:
Chapter 1:
1. Sec.1.2: Integrals as general and particular solutions; dierential equations
dy
= f (x) :
dx
2. Sec. 1.3: Direction (slope) elds, isoclines, solution curves. Ex
1
Lecture 39
= n2 and ( ) = a cos n + b sin n ; n = 0; 1; 2; : = 0 is also an eigenvalue).
Recall that
n
(recall that
Now we solve the equation r2 R00 + rR0 = R: I. = 0 ) r2 R00 + rR0 = 0 ) R00 = 0 Z R0 dR = R0 ln jR0 j = 1 ) r Z dr ) r ln r + A )
ln jR0
1
Lecture 38
=
2
III. Let
> 0. We have X 00 + 2 X = 0, whence X (x) = a cos x + b sin x:
X (0) = 0 ) a = 0, and X 0 (30) = 0 implies: b cos 30 = 0 ) 30 = ) So, and 60 (2n 1) x ; n = 1; 2; : : Xn (x) = sin 60
n
2n 2 1)
1 ; n = 1; 2; : :
=
(2n 60
=
(2n
1)
2