Math 334 Lecture #13
3.4: Reduction of Order; Repeated Roots
Method of Reduction of Order (dAlembert). Suppose y1 is a nonzero solution
of
y + p(t)y + q (t)y = 0.
For any scalar c, the scalar multiple cy1 is also a solution.
The idea of dAlembert to nd a
Problem 26 Page 143
I suppose you nished part (a) and part (b). Then from part (b), you should get ym
ym =
4(6 + )3
27(4 + )2
(1)
For part (c), we need to determine the smallest value of for which ym 4, hence
4(6 + )3
4 6 + 6 3
27(4 + )2
Hence the smalle
Here is the details: First dividing 4 on both sides and moving 27(4 + )2 on the right hand side:
4(6 + )3
4 (6 + )3 27(4 + )2
2
27(4 + )
(1)
Then we may do a sub u = 4 + , then we get
(6 + )3 27(4 + )2 (u + 2)3 27u2
(2)
then expand (u + 2)3 = u3 + 6u2 +
Here is the details: First dividing 4 on both sides and moving 27(4 + )2 on the right hand side:
4(6 + )3
4 (6 + )3 27(4 + )2
2
27(4 + )
(1)
Then we may do a sub u = 4 + , then we get
(6 + )3 27(4 + )2 (u + 2)3 27u2
(2)
then expand (u + 2)3 = u3 + 6u2 +
Problem 26 Page 143
I suppose you nished part (a) and part (b). Then from part (b), you should get ym
ym =
4(6 + )3
27(4 + )2
(1)
For part (c), we need to determine the smallest value of for which ym 4, hence
4(6 + )3
4 6 + 6 3
27(4 + )2
Hence the smalle
From part (a) and problem 18, we get
y (t) = ae
2t
3
2t
2
+ (1 + a)te 3
3
(1)
Part (b) is kind of trick. Here we dont need to take the derivative. When do you need to take the
derivative? Usually, when we want to nd the minimum or maximum value of y (t),
Suppose you get
x2
w=0
x1
Move the second term on the right hand side, then we get
w+
w =
x2
w
x1
divide w on both sides, you will get
w
x2
=
w
x1
and then integrate it on both sides
w
=
w
x2
x1
which is equivalent to
lnw =
x2
= x + ln(x 1)
x1
Then we can
Math 334 Lecture #26 6.4: Discontinuous Forcing Example. Use the Laplace transform to solve the IVP
y + 4y + 4y = u1 (t) u2 (t), y (0) = 0, y (0) = 0. Applying the Laplace tranform (by its rules) gives Lcfw_y + 4y + 4y = Lcfw_u1 (t) u2 (t) use linearity
Math 334 Lecture #25 6.3: Step Functions (and other Piecewise Continuous Functions) A Prototype. The unit step function with an upward step of 1 at c 0 is
uc (t) = [Sketch graph of the unit step function.] The Laplace transform of the unit step function i
Math 334 Lecture #24 6.2: Solving IVPs with the Laplace Transform Some Properties of the Laplace Transform.
(1) The Laplace transform is linear:
Lcfw_c1 f1 (t) + c2 f2 (t) =
0
est c1 f1 (t) + c2 f2 (t) dt
A
= lim
A
est c1 f1 (t) + c2 f2 (t) dt
0 A A
= lim
Math 334 Lecture #15
4.2: Homogeneous nth -order ODEs with Constant Coecients
A solution of the homogeneous linear ODE
L[y ] = a0 y (n) + a1 y (n1) + + an1 y + an y = 0
has the form y = ert if and only if r is a root of the characteristic equation
a0 rn +
Math 334 Lecture #16 3.5,4.3: Method of Undetermined Coecients
This is also known as the method of guessing as it applies to nth -order linear nonhomogeneous ODEs with constant coecients: L[y ] = an y (n) + an1 y (n1) + + a1 y + a0 y = g (t). The dierenti
Math 334 Lecture #17
3.6,4.4: Variation of Parameters
Theorem. If y1 , y2 , . . . , yn are linearly independent solutions of
L[y ] = y (n) + p1 (t)y (n1) + + pn1 (t)y + pn (t)y = 0,
then a particular solution of L[y ] = g (t) is
n
Yp =
ym (t)
m=1
g (t)Wm
Math 334 Lecture #18 3.7: Mechanical Vibrations Mass-Spring Systems. A spring is attached to a xed support, an object is suspended
on the spring, and a dashpot is placed about the object. [Sketch picture of this conguration.] Let u(t) be the position of t
Math 334 Lecture #19
3.8: Periodically Forced Vibrations
Undamped Periodically Forced Motion. This occurs when = 0:
mu + ku = F0 cos t,
where F0 is the forcing amplitude and is the forcing frequency.
[Recall that the natural frequency of undamped free mot
Math 334 Lecture #20
5.1: Power Series
Once a fundamental set of solutions of the associated homogeneous ODE for a linear
ODE with nonconstant coecients like
y + xy + 2y = g (x)
is found, then a particular solution is found by Variation of Parameters.
How
Math 334 Lecture #21
5.2: Power Series Solutions, Part I
Example. Solve the IVP
y + xy + 2y = 0,
y (0) = 4, y (0) = 1,
using a power series
an x n .
y=
n=0
[This is a mass-spring system with a varying damping constant. The center of the power
series guess
Math 334 Lecture #22
5.3: Power Series Solutions, Part II
Representation Principle. When can a general solution of
y + p(x)y + q (x)y = 0
be represented by power series?
If p(x) and q (x) are analytic at x0 , then the general solution of the ODE is
an (x
Math 334 Lecture #23 6.1: The Laplace Transform An Idea. Is there an invertible linear operator Lcfw_y (t) = Y (s) that tranforms an
IVP y (t) + ay (t) + by (t) = 0, y (0) = y0 , y (0) = y0 , into an algebraic equation like (s2 + as + b)Y (s) (as + b)y0 a
Math 334 Lecture #14 4.1: General Theory of nth -order Linear ODEs
An nth -order linear ODE has the form L [y ] = dn y dn1 y dy + p1 (t) n1 + + pn1 (t) + pn (t)y = g (t). n dt dt dt
[The linear dierential operator L acts on n-times dierentiable functions.
Yes, you are almost right. But remember that for two matrixes A,B , usually, we dont have
AB = BA.
Letg go back the question 7.4 number 6. We are trying to work with
x = Ax
we already get
x=
t t2
1 2t
x=
1 2t
02
And
And then we want to get A, where
A = x
Math 303 (Engineering Mathematics II)
Exam 1
RED KEY
Part I: Multiple Choice. Mark the correct answer on your scantron. Each question is worth 5
points. In problems 1 to ?, match the dierential equation to its direction eld.
1. y = y (y + 3)
Solution: d)
Math 303 (Engineering Mathematics II)
Exam 1
RED KEY
Part I: Multiple Choice. Mark the correct answer on your scantron. Each question is worth 5
points.
1. The solution to y =
a)
x2
is
y
3y 2 2x3 = C
d) x2 + y 3 = C
g)
b)
2x2 3y 3 = C
e) y 2 x3 = C
c) x2
F
Math 303 (Engineering Math II)
Exam 3
RED KEY
Part I: Multiple Choice Mark the correct answer on the bubble sheet provided.
1. What is the radius of convergence for
n=1
a)
0
b)
1
c)
2
d)
3
e)
4
f)
xn
?
n2
none of the above
Solution: The answer is (b).
2
1
1. Which dierential equation is equivalent to the system
x =x+y
y = x + 2y ?
a)
x x=0
b) x + 3x + x = 0
c) x 2x + x = 0
d)
x +x=0
e) x 3x + x = 0
f) x + 2x x = 0
Solution: e)
2. The Wronskian of the set of functions
1t
e,
1
2
e2t
1
is
c) 3et
3et
a)
3
b)