Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc
... the problem for u, (and
hence the problem for y).
As a check, then general solution for y is
y = −
1
3
cos 2x + c
1
cos x + c
2
sin x.
1115
We guess a particular solution of the form
y
p
= t
e
−t
(a ... y
2
=
e
3t
.
We compute the Wronskian of these solutions.
W (t) =
e
2t
e
3t
2
e
2t
3
e
3t
=
e
5t
We find a particular solution with variation of parameters.
y
p
= −
e...
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx
... cos
2
√
a
√
r
dr
1528
Hint 31 .17
Hint 31 .18
Hint 31 . 19
Hint 31 .20
Hint 31 .21
Hint 31 .22
Hint 31 . 23
1506
Exercise 31 .12
Find the inverse Laplace transform of
ˆ
f(s) =
1
s
3
− 2s
2
+ s − 2
with the following methods.
1. ... ı)
s=−ı
= ı
1
4
1 530
We substitute this into the third equation.
sˆy
3
= −ˆy
3
−
ˆy
3
s
−
ˆy
3
s
2
+
6
s
4
(s
3
+ s
2
+ s + 1)ˆy
3...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 9 ppt
... 7 .32
Hint 7 .33
Hint 7 .34
Hint 7 .35
Hint 7 .36
Hint 7 .37
30 0
Now to define the branch. We make a choice of angles.
z + 3 = r
1
e
ıθ
1
, −π < θ
1
< π
z = r
2
e
ıθ
2
, −
π
2
< θ
2
<
3
2
z ... z
1 /3
=
3
√
z
e
ık2π /3
, k = 0, 1, 2
2.
g(w) = w
3
= |w|
3
e
3 arg(w)
32 8
3
1+ı
=
e
(1+ı) log 3
=
e
(1+ı)(ln 3+ ı2πn)
=
e
ln 3 2πn
e
ı(ln 3+ 2πn)
, n ∈ Z
√...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
... =
3 −
√
2
1 /3
3 +
√
2
1 /3
( 3 + 2)
1 /3
=
3
3 +
√
2
e
ıπ /3
3
3 −
√
2
e
ıπ /3
3
√
1
e
ıπ /3
=
3
√
7
e
ıπ
= −
3
√
7
The value of the function is
w =
3
√
abc
e
ı(α+β+γ) /3
.
Consider the ... real variable counterpart.
36 4
With
f(z) = z
1 /3
(z −ı)
−1 /3
(z + ı)
−1 /3
=
3
√
r
e
ıθ /3
1
3
√
s
e
−ıφ /3
1
3
√
t
e
−ıψ /3
=
3
r
st
e
ı(θ−φ−ψ) /3...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... for the coefficients.
a + b + c = 2
4a + 2b + c = 6
9a + 3b + c = 12
p(n) = n
2
+ n
We e xamine the first few partial sums.
S
1
=
1
2
S
2
=
2
3
S
3
=
3
4
S
4
=
4
5
592
5.
∞
n=1
ln (2
n
)
ln (3
n
) ... 12. 13
CONTINUE.
1.
∞
n=0
z
n
(z + 3)
n
2.
∞
n=2
Log z
ln n
3.
∞
n=1
z
n
4.
∞
n=1
(z + 2)
2
n
2
594
Thus the series converges absolutely for |z| < 1/4.
6. By the Ca...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt
... integral converges for (a) < 1.
725
Figure 13. 8: The real and imaginary part of the integrand for several values of α.
Note that the integral exists for all nonzero real α and that
lim
α→0
+
1
−1
1
x ... = ı
= ıπ
3
lim
z→ı
1
z + ı
=
π
3
2
Now we return to Equation 13. 1.
2
∞
0
ln
2
x
1 + x
2
dx + ı2π
∞
0
ln x
1 + x
2
dx =
π
3
4
We equate the real and imaginary...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx
... x
n
F
1,
y
x
.
(Just formally substitute 1/x for λ.) For example,
xy
2
,
x
2
y + 2y
3
x + y
, x cos(y/x)
are homogeneous functions of orders 3, 2 and 1, respectively.
Euler’s theorem for a homogeneous ... αy(t).
775
1
2
3
4
4
8
12
16
1
2
3
4
4
8
12
16
Figure 14.1: The p opulation of bacteria.
1
2
3
4
32
64
96
128
1
2
3
4
32
64
96
128
Figure 14.2: The discrete po...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 2 potx
... equation
dw
dz
+
1
z
2
+ 9
w = 0.
We factor the denominator of the fraction to see that z = 3 and z = − 3 are regular singular points.
dw
dz
+
1
(z − 3) (z + 3)
w = 0
We make the transformation z = 1/ζ ... rearranging terms to form exact derivatives.
4yy
− xy
− y + 1 −9x
2
= 0
d
dx
2y
2
− xy
+ 1 − 9x
2
= 0
2y
2
− xy + x −3x
3
+ c = 0
y =
1
4
x ±
x
2
− 8(c + x − 3x
3
)
...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx
... of η to be zero.
4 3 −2
8 −6 −4
−4 3 2
0
0
η
3
= c
1
1
0
2
+ c
2
0
2
3
−2η
3
= c
1
, −4η
3
= 2c
2
, 2η
3
= 2c
1
− 3c
2
c
1
= c
2
, η
3
= −
c
1
2
888
We see that ... 0
e
2
e
2
e
2
/2
0
e
2
e
2
0 0
e
2
0 3 −2
0 −1 −1
1 −1 −1
e
A
=
0 2 2
3 1 −1
−5 3 5
e
2
2
8 59
Thus there are two eigenvectors.
4 3 −2
8 −6 −4
−...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot
... =
e
x
e
2x
e
x
2
e
2x
= 2
e
3x
−
e
3x
=
e
3x
.
91 0
Consider L[y] = p
n
y
(n)
+ ···+ p
0
y. If each of the p
k
is k times continuously differentiable and u and v are n times
continuously differentiable ... 0
¯y
(n)
+ p
n−1
¯y
(n−1)
+ ··· + p
0
¯y = 0
L [¯y] = 0
9 03
16 .3 Transformation to a First Order System
Any linear differential equation can be put in the form of a...
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