Unformatted text preview: 5t . Now, to ﬁnd a particular solution, we assume that it can be written in the form 1
2
x p = u1
+ u2
e 5t .
2
1
As shown in the course notes, this requires us to solve the system
)= 1
t (2) )=4+ 2
t (3) u01 (1) + u02 ( 2e
u01 (2) + u02 (e 5t 5t , AMATH 350 Page 3 Assignment #7 Solutions  Fall 2013 Multiplying (3) by 2, and then subtracting it from (4), we obtain
5e 5t 0
u2 4
Therefore u02 = e5t , so
5
u2 =
From (3) we also know that u01 = = 4. 4 5t
e.
25 1
+ 2e
t 5t 0
u2 u1 = ln t + = 18
+ , so
t5 8t
.
5 Therefore a particular solution is
✓
◆ 8t
4
1
2
xp = ln t +
+
2
1
5
25 8
8
ln t + 5t 25
=
,
16t
4
2 ln t + 5 + 25
and the full general solution is 1
2
x = C1
+ C2
e
2
1 5t ✓ 8t
+ ln t +
5 ◆ 1
2 4
+
25 2
1 . AMATH 350 Page 4 Assignment #7 Solutions  Fall 2013 3. This is simply a diﬀerent way of expressing the same kinds of problems we’ve
been solving; we have 11
1
0
0
~=
x
~+
x
2t,
~ (0) =
x
.
11
1
0
The characteristic equation is 1
1
det
=(
1
1 1) 2 1= 2 2= ( 2) = 0 , so the eigenvalues are 0 and 2.
To ﬁnd an eigenvector for 0, we write 11
v1
11
v2 = 0
0 and conclude that v1 + v2 = 0, so we may use ~ =
v 1
.
1 For the eigenvector 2, we have instead 11
v1
0
=
1
1
v2
0 1
and so v1 v2 = 0, and we may use ~ =
v
.
1 We can now see that the complementary function is 1
1
~ h = c1
x
+ c2
e 2t .
1
1
Therefore, to ﬁnd the general solution, we let 1
1
~ = u1
x
+ u2
1
1 e 2t and to ﬁnd u1 and u2 we must solve the system of equations
u01 + u02 e2t = 2t
u01 + u02 = 2t. Adding these, we ﬁnd that
2u02 e2t = 0 =) u02 = 0 =) u2 = c2 .
Subtracting them instead gives us
2u01 = 4t =) u01 = 2t =) u1 = t2 + c1 .
Thus we’ve found that
~ = c1
x 1
1 + c2 1
1 2t e+ 1
1 t2 . Applying the initial condition, we require that
c1 + c2 = 0
c1 + c2 = 0,
so c1 = c2 = 0, and the desired solution is simply x (t) = t2 , y (t) = t2 . AMATH 350 Page 5 Assignment #7 Solutions  Fall 2013 4. Commodity Problem:
7. (a) The model is
dD
= 2D 5p + 10,
dt
dp
= D S = D 50,
dt
or y0 = Ay + F where
D
p y=...
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 Fall '12
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