1. a.
X
0
+
λX
=0
X
(0) = 0
X
(
π
)=0
Try
e
rx
. As we know from ODEs, this leads to the characteristic equation for
r
r
2
+
λ
=0
Or
r
=
±
√
−
λ
We now consider three cases depending on the sign of
λ
Case 1:
λ<
0
In this case
r
is the square root of a positive
number and thus we have two real roots. In
this case the solution is a linear combination of two real exponentials
X
=
A
1
e
√
−
λx
+
B
1
e
−
√
−
λx
It is well known that the solution can also be written as a combination of hyperbolic sine
and cosine, i.e.
X
=
A
2
cosh
√
−
λx
+
B
2
sinh
√
−
λx
The other two forms are may be less known, but easily proven. The solution can be written
as a shifted hyperbolic cosine (sine). The proof is straight forward by using the formula for
cosh(
a
+
b
)(s
inh(
a
+
b
))
X
=
A
3
cosh
√
−
λx
+
B
3
Or
X
=
A
4
sinh
√
−
λx
+
B
4
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 BELL,D
 Complex number, boundary condition, hyperbolic cosine, hypebolic sine, shifted hyperbolic cosine

Click to edit the document details