Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
... y
1/2
exp
−
2y
3/ 2
3
x exp
−
2y
3/ 2
3
= −exp
−
2y
3/ 2
3
+ c
1
x = −1 + c
1
exp
2y
3/ 2
3
x + 1
c
1
= exp
2y
3/ 2
3
log
x + 1
c
1
=
2
3
y
3/ 2
y =
3
2
log
x + 1
c
1
2 /3
y =
c +
3
2
log(x ... 3p
)u
exp
−
1
3
p(x) dx
y
=
u
− pu
+
1
3
(p
2
− 3p
)u
+
1
27
(9p
− 9p
− p
3
)u
exp
−
1
3
p(x) dx
...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx
... x
=
dx
dy
.
dy
dx
=
1
y
3
− xy
2
dx
dy
= y
3
− xy
2
x
+ y
2
x = y
3
Now we have a first order equation for x.
d
dy
e
y
3
/3
x
= y
3
e
y
3
/3
x =
e
−y
3
/3
y
3
e
y
3
/3
dy + c
e
−y
3
/3
Example 18 .3. 2 Consider ... equation of order
n − 1 for u
. Writing the derivatives of
e
u(x)
,
d
dx
e
u
= u
e
u
d
2
dx
2
e
u
= (u
+ (u
)
2
)
e
u
d
3
dx
3
e...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx
... 1)
1
3
x
3
+ x
1
3
−
1
2
−
1
3
x
3
+
1
2
x
2
y =
1
6
(x
3
− x).
Example 21 .7. 4 Find the solution to the differential equation
y
− y = sin x,
that is bounded for all x.
The Green function for ... the homogeneous solutions to y
p
and it will still be a particular solution. For example,
η
p
= −
1
3
sin(2x) −
1
3
sin x
= −
2
3
sin
3x
2
cos
x
2
is a partic...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
... ı
2
n
by
successive squaring.
√
3 + ı
2
= 2 + ı2
√
3
√
3 + ı
4
= −8 + ı8
√
3
√
3 + ı
8
= −128 − ı128
√
3
√
3 + ı
16
= 32 76 8 + 32 76 8
√
3
Next we multiply
√
3 + ı
4
and
√
3 + ı
16
to obtain ... answer.
√
3 + ı
20
=
32 76 8 + 32 76 8
√
3
−8 + ı8
√
3
= −524288 − ı524288
√
3
Since we know that arctan
√
3, 1
= π/6, it...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt
... ı2
√
3
−2 + ı2
√
3
4
−1
=
−2 + ı2
√
3
−8 − ı8
√
3
2
−1
=
−2 + ı2
√
3
−128 + ı128
√
3
−1
=
−512 − ı512
√
3
−1
=
1
512
−1
1 + ı
√
3
=
1
512
−1
1 + ı
√
3
1 − ı
√
3
1 ... u
2
3
v =
(u
0
w
0
+ u
1
w
1
+ u
2
w
2
+ u
3
w
3
) + ı (−u
1
w
0
+ u
0
w
1
+ u
3
w
2
− u
2
w
3
) + (−u
2
w
0
− u
3
w
1
+ u
0
w
2
+ u
1
w
3
)
+ k (−u
3
w
0
+...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... ∆r cos θ and y = ∆r sin θ.
f
(0) = lim
∆r→0
f
∆r
e
ıθ
∆r
e
ıθ
= lim
∆r→0
∆r
4 /3
cos
4 /3
θ∆r
5 /3
sin
5 /3
θ+ı∆r
5 /3
cos
5 /3
θ∆r
4 /3
sin
4 /3
θ
∆r
2
∆r
e
ıθ
= lim
∆r→0
cos
4 /3
θ sin
5 /3
θ + ı ... function
f(z) = u + ıv =
x
3
(1+ı)−y
3
(1−ı)
x
2
+y
2
for z = 0,
0 for z = 0.
Show that the partial derivatives of u and v with respect to x and y exist at z =...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... integrand to s ee that there are singularities at the cube roots of 9.
z
z
3
− 9
=
z
z −
3
√
9
z −
3
√
9
e
ı2π /3
z −
3
√
9
e
−ı2π /3
Let C
1
, C
2
and C
3
be contours around z =
3
√
9, ... Integral Formula to evaluate the integrals along C
1
and C
2
.
C
(z
3
+ z + ı) sin z
z
4
+ ız
3
dz =
C
1
(z
3
+ z + ı) sin z
z
3
(z + ı)
dz +
C
2
(z
3
+ z + ı) s...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
... ,
converges for α > 1 and diverges for α ≤ 1.
Hint, Solution
564
Example 12 .3. 2 Convergence and Uniform Convergence. Consider the series
log(1 − z) = −
∞
n=1
z
n
n
.
This series converges for |z| ... expansions of 1/(1 + z).
For |z| < 1,
1
1 + z
= 1 +
−1
1
z +
−1
2
z
2
+
−1
3
z
3
+ ···
= 1 + (−1)
1
z + (−1)
2
z
2
+ (−1)
3
z
3
+ ···
= 1 − z + z
2
− z...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf
... 1))
z
n
+ d, for 2 < |z|
Solution 12.29
The radius of convergence of the series for f(z) is
R = lim
n→∞
k
3
/3
k
(k + 1)
3
/3
k+1
= 3 lim
n→∞
k
3
(k + 1)
3
= 3.
622
The ... 13. 1 .3 Evaluate the integral
C
cot z coth z
z
3
dz
where C is the unit circle about the origin in the positive direction.
The integrand is
cot z coth z
z
3
=...
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 ... 0
y
−
− y
−
= 0, y
−
(0) = y
+
(0), for x < 0,
and define the solution, y, to be
y(x) =
y
+
(x), for x ≥ 0,
y
−
(x), for x ≤ 0.
The initial condition for y
−
d...
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