Solutions to Problem Set 1
ARE 261
Question 1
Begin with the assumption that the solution to the second order differential equation is of the form
y
=
k
exp
xt
, where
k
and
x
are constants. Then ˙
y
=
kx
exp
xt
and ¨
y
=
kx
2
exp
xt
. Substitute these
expressions for
y,
˙
y
and ¨
y
into the differential equation and divide through by
k
exp
xt
to get
x
2
+
ax
+
b
= 0
(1)
This gives a quadratic in x which we assume has two distinct root,
x
1
and
x
2
given by
x
1
=

a
2
+
√
a
2

4
b
2
(2)
and
x
2
=

a
2

√
a
2

4
b
2
(3)
The solution to the second order differential equation is then given by
y
=
k
1
exp
x
1
t
+
k
2
exp
x
2
t
(4)
where
k
1
and
k
2
are constants to be determined by boundary or initial conditions.
In the next
subpart we are given two such initial conditions,
y
(0) = 0 and ˙
y
(0) = 1. Substitute these conditions
1
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into equation (4) and its first derivative with respect to time to get the following two equations
k
1
+
k
2
= 0
(5)
k
1
x
1
+
k
2
x
2
= 0
(6)
These can be solved for
k
1
and
k
2
to get the complete solution for the differential equation. The
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 Fall '06
 KARP
 Derivative, Expression, order differential equation

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