CME 102, Winter 2009
Ordinary Differential Equations for Engineers
Prof. Eric Darve
Homework 1 – Solutions
1. Kreyszig:
(a) exercise 6 page 8
This is a second order differential equation. We take twice the derivative of
y
to
get
y
00
=

aπ
2
cos(
πx
)

bπ
2
sin(
πx
) =

π
2
[
a
cos(
πx
) +
b
sin(
πx
)] =

π
2
y.
We can thus conclude that
y
00
+
π
2
y
= 0.
(b) exercise 16 page 8
Substitution of
y
l
=
cx

c
2
into the ODE gives
(
y
0
l
)
2

xy
0
l
+
y
l
=
c
2

xc
+ (
xc

c
2
) = 0
thus proving that
y
l
=
cx

c
2
is the general solution.
Similarly, for
y
p
=
1
4
x
2
, we get that
(
y
0
p
)
2

xy
0
p
+
y
p
=
2
4
x
2

x
2
4
x
+
1
4
x
2
=
1
4

1
2
+
1
4
x
2
= 0
.
In figure 6, we can see that the parabola represents the function
y
p
. Such parabola
is the
envelope
of the set of general solutions, the lines
y
l
(recall that
c
is an
arbitrary constant).
For a given
c
, the corresponding line
y
l
is tangent to the
parabola
y
p
at exactly one point.
(c) exercise 20 page 8
From the physical information, we have that the pressure satisfies an ODE of the
form
y
0
=
ky
. The general solution to this ODE is
y
=
y
0
exp(
kx
), where
y
0
is
the pressure at sea level
x
= 0. Now we also know that
y
(18000) =
1
2
y
0
=
y
0
exp(
k.
18000)
.
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 Winter '09
 Darve,E
 Derivative, Prof. Eric Darve

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