Unformatted text preview: on is
y (0) = 1000.
The DE is linear, and an integrating factor is
µ=e R 1
dt
20 t/20 =e . Factoring this into the equation, we obtain
e t/20 dy 1
e
20 dt t/20 y = 365e t/20 which must be equivalent to
d
e
dt t/20 = 365e t/20 (and we can quickly check that it is, so we have calculated our integrating
factor correctly).
Integrating, we ﬁnd
e
so t/20 y= t/20 7300e y= +C 7300 + Cet/20 . Apply the initial condition:
=) y (0) = 1000 =) =) 1000 = 7300 + C C = 8300
y = 8300et/20 7300. Therefore, after 10 years, the account balance should be approximately
y (10) = 8300e1/2
= $6384.39. 7300 AMATH 350 Page 5 Assignment #3  Fall 2013 4. (Consider a commodity (such as wheat) where the supply ﬂuctuates on an
annual basis, i.e., S = 5(1 + cos(2⇡ t)), where t is measured in years. Suppose
that the rate of increase of price is equal to the rate of decrease of quantity
and the initial price is p0 . )
a) (If the demand for this commodity is given by D = 20 p/10, ﬁnd the
price as a function of time. What is the long term behaviour of the price?
How does this depend on p0 ? Are there any equilibrium solutions? Use
Maple or Matlab to sketch the solution for diﬀerent values of p0 . )
Solution:
dp
= D S = 20 p/10 5(1 + cos(2⇡ t)), p(0) = p0 .
dt
1
dp
+ p = 15 5 cos(2⇡ t).
Linear DE with standard form:
dt 10
Integrating factor: e 1/10 dt = et/10
dp
= D S = 20 p/10 5(1 + cos(2⇡ t)), p(0) = p0 .
5. (a)
dt
dp et/10
t/10
1
et/10 +
= 15edp
5et/10 cos(2⇡ t)
+ p = 15 5 cos(2⇡ t).
Linear DE with standard form:
dt
10p
dt 10
d e1
Integrating factor: t/10 p/10 dt =15t/1010 5et/10 cos(2⇡ t)
[e = e et/
dt t/10
e
t/10 dp
Integrate by parts twice
e
+
= 15et/10 5et/10 cos(2⇡ t) on RHS (details below)
dt
10p
50
t/10
et/10
et/10 cos(2⇡ t) + 20⇡ et/10 sin(2⇡ t) + C
d e 10 p = 150t/10
t/
1t/10400⇡ 2⇡ t)
+ cos(2
[e p] = 15e
5e
dt
50
t
[cos(2⇡ t + RHS (details below)/10
General Solution: p(t) = 150
Integrate 400parts twice )on 20⇡ sin(2⇡ t)] + Ce
1 + by ⇡ 2
50
et/10 p = 150et/10
et/10 cos(2⇡50) + 20⇡ et/10 sin(2⇡ t) + C
t
50
IC: p(0) = p0 ) p0 = 150 1+400 2 + C 1 +C =⇡ 20 150 + 1+400 2
) 400 p
50
50
50
[cos(2⇡ t + 20⇡ sin(2 0 150+ t/10
General Solution: (t =
Solution of IVP: p(t)p= )150 150 1 2 [cos(22 t) + 20⇡)sin(2⇡ t)]+(p⇡...
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This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.
 Fall '12
 DavidHamsworth

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