Unformatted text preview: n integrating factor is e 2dx = e2x . The equation
dx
d 2x
can be rewritten as
e y = ex , and so we ﬁnd that
dx a) For y=e 2x (ex + C ) = e x + Ce 2x . We can see that y = e x is an exceptional solution, and all other solutions
approach it as x ! 1, and are repelled by it as x ! 1. AMATH 350 Page 3 Assignment #3 Solutions Next, observe that y 0 = 0 if and only if y = 1 e x (this is our horizontal
2
isocline). The solutions must look something like this:
3 2 1 5 4 3 2 1 C>0
← C=0 0 1 C<0
2 Figure 1:
dy
b) For x
= 3x 2
dx
form: 1 2 3 4 5 ↖ All solutions with C<0
eventually cross the xaxis,
increasing until they cross
the horizontal isocline,
and then decrease alongside
the exceptional solution. 3 y we need to start by putting the equation in standard
dy 1
+ y = 3x.
dx x
R 1 An integrating factor is µ(x) = e x dx = eln x = x. Incorporating this, we
obtain
dy
x + y = 3x2
dx
which must be the same equation as
dy
(xy ) = 3x2 .
dx
Therefore xy = x3 + C and so C
.
x
Now, y = x2 is an exceptional solution. All other solutions approach this
as x ! 1 and diverge from it as x ! 0 every solution has a vertical
asymptote at the origin (except for the exceptional solution).
The horizontal isocline is y = 3x2 . Finally, notice that C > 0 makes
y > x2 if x > 0, but y < x2 if x < 0.
Putting our observations together, we arrive at the following sketch:
y = x2 + AMATH 350 Page 4 Assignment #3 Solutions 3 C<0 C>0
2 1 ← C=0
5 4 3 2 1 0 C>0 1 1 2 3 4 5 C<0 2 Figure 2: 3 3. (Interest Rate Problem)
Let’s let y be the balance of the account, and measure t in years. For continuously compounded interest alone, the balance would evolve according to the
dy
Mathusian model:
= 0.05y . To account for continuous depositing at the
dt
rate of $365 per annum, we simply add a term:
dy
= 0.05y + 365.
dt
The initial condition 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...
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 Fall '12
 DavidHamsworth
 Cos, Boundary value problem, dt

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