Unformatted text preview: 1/2 ) v 0 = 1 p 3/2 p0 . Using the
Let v =
=
2
2
18
1 + Ce t/2
DE:
1 = 1 + C1 e t/2 =
vv +
=
4
1 8 4/2 Linear DE
1
0
v =2
p1 +
16 2
8
Standard form: v 0 p 1 v v 8
factor:
General solution:+ 2= = 12. =Integrating t12 2 1e 1/2 dt = et/2 .
/)
(1 + Ce v +
Linear DE
=
1 t/2 16
1 t/2 8 4
2e
t/2 0
)
1.
Apply IC: p(0) = p0 e) v 0+ 2 e v = 2 8 C =
p=
p0 1/2 dt
1
1 (1 + C )
0
Standard form: v + 2 v = d . t/2
Integrating 2factor: e
= et/2 .
1 t/
8
[e v ] 16
=
e
Solution of IVP: p(t) = dt
8 2.
e
(1 + ( 14 t/ 1)1 t/1)2
p
v
eC
et/2 v 0 + t/2e0 2= =et/2 +t/2
e2 v
1
48
(b) Note that limt
p(t) = 16.
d t/2
1 t/2
t/2
1
If the initial price is greater than $16 the pricet/will 1 + Ce with limiting value $16.
[ev v ] = + Cee 2 = decrease
=
1
dt
48
If the initial price is equal to $16, it will remain at $16. 4
1 t/2
16 v price ewill+ C1
If the initial price is less than $16 2the =
increase with limiting value $16.
et/
General solution: p = v 2 =
4
t/2 )2
(1 + Ce
14
1 + Ce t/2
16
v C = + C11. t/2 =
e
)=
Apply IC: p(0) = p0 ) p0 =
(1 + C )2
4
4 4 p0
16
16
Solution of IVP: p(t) =
.
General solution: p = (1 + = 4 0 1)e t/t2/)2 2
v 2( p
(1 + Ce 2 )
(b) Note that limt
p(t) = 16.
4
16
b) (Discuss the(0) =is 0 ) p0 of your ﬁndings for the application. That is, $16.
implications =
)C=
1
Applyinitial price pgreater than $16 the price will decrease .with limiting value
If the IC: p
2
p0
describe the price is equal to $16, it will remainand how this depends on p0 .
behaviour of the(1 + Cin time at $16.
price )
If the initial
You need of IVP: is(less than $16 ) 16
If the initial price p t) =
the
increase with limiting value $16.
Solution only consider p0 > 0. 4 price will 2 .
(1 + ( p0 1)e t/2 )
Solution:
(b) Note that limt
p(t) = 16.
4
If the initial price is greater than $16 the price will decrease with limiting value $16.
If the initial price is equal to $16, it will remain at $16.
If the initial price is less than $16 the price will increase with limiting value $16. 4...
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 Fall '11
 Noidea
 Cos, Boundary value problem, dt

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