Unformatted text preview: we ﬁnd that C = 0, so y = 2 tan (2x) 4x.
(Of course, since the tangent function has period ⇡ , we could use any
multiple of ⇡ for C ; we’d still have the same function.)
= 2y + e x an integrating factor is e 2dx = e2x . The equation
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 - Fall 2013 Next, observe that y 0 = 0 if and only if y = 1 e x (this is our horizontal
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:
b) For x
= 3x 2
form: 1 2 3 4 5 ↖ All solutions with C<0
eventually cross the x-axis,
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
+ y = 3x.
R 1 An integrating factor is µ(x) = e x dx = eln x = x. Incorporating this, we
x + y = 3x2
which must be the same equation as
(xy ) = 3x2 .
Therefore xy = x3 + C and so C
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 - Fall 2013 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
= 0.05y . To account for continuous depositing at the
rate of $365 per annum, we simply add a term:
= 0.05y + 365.
The initial conditi...
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- Cos, Boundary value problem, dt