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Unformatted text preview: 28 Economic Dynamics
Example 2.1
Consider (iii) above. This is an autonomous ﬁrstorder homogeneous differential
equation. Rearranging the equation we have
dx 1
=k
dt x
Integrating both sides with respect to t yields
dx 1
dt =
dt x k dt ln x(t) = kt + c0
where c0 is the constant of integration. Taking exponentials of both sides yields
x(t) = cekt
where c = ec0 . It is readily veriﬁed that this is indeed a solution by differentiating
it and substituting. Thus
kcekt = kx = kcekt
which holds identically for any a < t < b.
Example 2.2
To check whether x(t) = 1 + t + cet is a solution of dx/dt = x − t, we can differentiate x with respect to t and check whether the differential equation holds exactly.
Thus
dx
= 1 + cet
dt
... 1 + cet = 1 + t + cet − t
Hence x(t) = 1 + t + cet is indeed a solution.
Example 2.3
Check whether
p(t) = ap0
bp0 + (a − bp0 )e−at is a solution to the differential equation
dp
= p(a − bp)
dt
Differentiating the solution function with respect to t we obtain
dp
= −ap0 [bp0 + (a − bp0 )e−at ]−2 (−a(a − bp0 )e−at )
dt
a2 p0 (a − bp0 )e−at
=
[bp0 + (a − bp0 )e−at ]2 ...
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 Fall '11
 Dr.Gwartney
 Economics

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