Unformatted text preview: + Dxe +8Ax + 4B + 4De x x 4Dxe +3Ax2 + 3Bx + 3C + 3Dxe x x = x2 + e x . Matching coeﬃcients, we have
3A = 1
8A + 3B = 0
2A + 4 B + 3C = 0
2D = 1
and solving these yields A = 1/3, B = 8/9, C = 26/27, and D = 1/2.
Putting everything together, the solution to the given ODE is
y = C1 e 3x + C2 e x 1
+ x2
3 8
26 1
x+
+ xe x .
9
27 2 AMATH 350 3. Page 3 Assignment #4 Solutions Winter 2013 a) Show that the price satisﬁes the secondorder DE
d2 p
dp
+ 8 + 25p = 100.
2
dt
dt
dq
dS
dp
=
= D S with D = 10 8p and
= 25p
4. (a) We have
dt
dt
dt
ﬁrst equation and then using the second gives 100. Di erentiating the dp
d2 p
dp
d2dp dDdq dS
p
=
= S8 with D5p + 100 8p and dS = 25p 8 100+Di pe= 100.
2 = 10
or
+
25 rentiating the
4. (a) We have 2 = dt = D
2
dtdt
dt
dt .
dt dt
dt dt
ﬁrst equation and then using the second gives
00
(b) If the initial value of the price ispp0+and + 25p = 0. derivative of the price
8p0 the initial
b) Associated homogeneous equation:
2
8+ 6
2
d2
dp How does
d (t . What is8 + 25 0 ) behaviour 2 4 100p price? 3
Auxili...
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 Winter '14
 Heterogeneity, Homogeneity, yp

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