Unformatted text preview: Continuous dynamic systems 37
Figure 2.5. which satisﬁes the initial condition
1
v0 = k0−α This allows us to solve for k(t) as follows
k1−α =
i.e. as
as
1
+ k0−α −
e−(1−α)(n+δ)t
n+δ
n+δ k(t) = as
as
1
+ e−(1−α)(n+δ)t k0−α −
n+δ
n+δ 1
1−α The solution path for different initial values of k is illustrated in ﬁgure 2.5. 2.2 Solutions to ﬁrstorder linear differential equations
Solutions to ﬁrstorder linear differential equations are well discussed in the mathematical texts on differential equations (see Boyce and DiPrima 1997; Giordano
and Weir 1991). Here our intention is simply to provide the steps in obtaining a
solution. In doing this we shall suppose y is a function of t. This is useful since
most economic examples are of this type. The general form for a ﬁrstorder linear
differential equation is then
dy
+ p(t)y = g(t)
dt
Notice that in this formulation both p and g are functions of time. This also allows
for the case where p(t) and g(t) are constants, in which case we have
dy
+ by = a
dt
The fourstep procedure is as follows.
Step 1 Write the linear ﬁrstorder equation in the standard form
dy
+ p(t)y = g(t)
dt ...
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 Fall '11
 Dr.Gwartney
 Economics, Derivative, Linear Differential Equations, Giordano, Boyce, Weir

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