... transform of ˆy(s) by first finding its partial fraction expansion.
ˆy(s) =
s/3
s
2
+ 1
−
s/3
s
2
+ 4
+
s
s
2
+ 1
= −
s/3
s
2
+ 4
+
4s/3
s
2
+ 1
y(t) = −
1
3
cos(2t) +
4
3
cos(t)
Example 31 .4. 3 ... transform methods and show that
i
1
=
E
0
2R
+
E
0
2ωL
e
−αt
sin(ωt)
where
α =
R
2L
and ω
2
=
2
LC
− α
2
.
Hint, Solution
Exercise 31.23
Solve the initial value problem,
y
+ 4y
+ 4...
... 1)c
n+2
+ c
n−2
= 0, for n ≥ 2
c
n +4
= −
c
n
(n + 4) (n + 3)
For our first solution we have the difference equation
a
0
= 1, a
1
= 0, a
2
= 0, a
3
= 0, a
n +4
= −
a
n
(n + 4) (n + 3)
.
For our second solution,
b
0
= ... polynomial are
α
1
=
1 +
1 − 3 /4
2
=
3
4
, α
2
=
1 −
1 − 3 /4
2
=
1
4
.
Thus our two series solutions will be of the form
w
1
= z
3 /4
∞
n=0
a
n
z
n
, w...
... =
1
6
−
1
π
2
∞
n=1
1
n
2
∞
n=1
1
n
2
=
π
2
6
141 8
29. 5 Hints
Hint 29. 1
Hint 29. 2
Hint 29. 3
Hint 29 .4
Write the problem in Sturm-Liouville form to show that the eigenfunctions are orthogonal ... πx)
2
dx
=
π
3
/3
π
5
/30
=
10
π
2
≈ 1.013
142 9
29 .4 Exercises
Exercise 29. 1
Find the eigenvalues and eigenfunctions of
y
+ 2αy
+ λy = 0, y(a) = y(b) = 0,
where a < b....
... 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 ... v +
b
a
G(x|ξ)f(ξ) dξ.
1 099
A particular solution is
y
p
=
t
3
6
e
t
+4.
The general solution of the differential equation is
y = c
1
e
t
+c
2
t
e
t
+
t
3
6
e
t
+4.
We use the in...
... x
3
, x
4
} is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo-
12 84
1
2
3
4
5
6
-60
-40
-20
Figure 24. 3: log(error in approximation)
In Figure 24. 4 we see ... 21)
π
4
P
4
(x)
=
105
8π
4
[(315 − 30π
2
)x
4
+ ( 24
2
− 270)x
2
+ (27 − 2π
2
)]
The cosine and this polynomial are plotted in the second graph in Figure 25.1. The le ast square...
... the
formula to obtain information about the eigenvalues before we solve a problem.
Example 27 .4. 2 Consider the self-adjoint eigenvalue problem
−y
= λy, y(0) = y(π) = 0.
1322
Example 27 .4. 1 ... eigenfunction
φ. Green’s formula states
φ|L[φ] − L[φ]|φ = 0
φ|λφ − λφ|φ = 0
(λ − λ)φ|φ = 0
Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real.
13 19
27.7 Hints
Hint 27.1
1327
2. F...
... solution and discuss in as much detail as possible what goes wrong.
13 64
Note that this formula is valid for m = 0, 1, 2, . .
Similarly, we can multiply by sin(mx) and integrate to solve for b
m
. ... ∼
∞
n=1
oddn
4
n
x
−π
sin(nt) dt
=
∞
n=1
oddn
4
n
−
1
n
cos(nt)
x
−π
=
∞
n=1
oddn
4
n
2
(−cos(nx) + (−1)
n
)
= 4
∞
n=1
oddn
−1
n
2
− 4
∞
n=1
oddn
cos(nx)
n
2...