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Midterm Exam
Math S3027 - T. Collins
Solve all 5 problems.
1. (10 points) Let a 6= 0, −1 be a non-zero, real number. Find the general solution
of the initial value problem
ty 0 + ay = t1−a sin(t) + ta+2 ,
t > 0.
Solution:
Use the method of integrating factors. Divide the equation by t. Let µ(t) be the
integrating factor, which must satisfy µ0 = at−1 µ. We find µ(t) = ea ln(t) = ta .
The equation is then (ta y)0 = sin(t) + t2a+1 . Since a 6= −1 we know 2a + 1 6= −1
so we can integrate the equation to obtain
y(t) = −t−a cos(t) +
1
t−a+2 + Ct−a .
2a + 2
2. Let a ∈ R be any real number. Consider the initial value problem:
y 00 − 2y 0 − 15y = 0,
y(0) = a, y 0 (0) = a2 .
(a) (10 points) Solve the initial value problem.
Solution:
The characteristic equation is r2 − 2r − 15 = 0, or (r − 5)(r + 3) = 0. As a
result, we find y1 = e5t and y2 = e−3t . Then y = C1 y1 + C2 y2 . Plugging in the
initial condition we obtain C1 + C2 = a and 5C1 − 3C2 = a2 . Solve this system
2
2
, and C2 = 5a−a
.
to obtain C1 = 3a+a
8
8
(b) (5 points) For what values of a, does the solution approach −∞ as t → +∞?
For what values of a does the solution remain bounded as t → +∞.
Solution:
Since e−3t is bounded as t → +∞, and e45 → ∞ as t → +∞, we need to determine the sign of C1 . If C1 > 0, then y → +∞, and if C1 < 0 then y → −∞,
and if C1 = 0 then y(t) is bounded. Thus, y(t) goes to −∞ if and only if
3a + a2 < 0, which is achieved for −3 < a < 0. The solution is bounded if and
only if a = −3, 0.
3. (a) (10 points) Consider the differential equation
ty 00 − (t + 1)y 0 + y = 0
t > 0.
Given that y1 (t) = 1 + t is a solution, find a second, linearly independent soluet
tet
tion. (Hint: Observe that dtd ( 1+t
) = (1+t)
2 ).
Solution:
Use the method of reduction of order. We guess y2 (t) = u(t)(1 + t). Plug this
into the equation to find that y2 (t) is a solution provided
t(1 + t)u00 − (1 + t)2 − 2t u0 = 0.
In other words, we have
00
u =
t
1
+
t(1 + t) 1 + t
u0 .
Using partial fractions, this equation can be written as
log(u0 )0 = 1 +
Integrate to find u0 =
tet
.
(1+t)2
1
2
−
.
t 1+t
By the hint, u =
et
.
1+t
Thus, y2 (t) = et .
(b) (10 points) Find the general solution of the inhomogeneous equation
ty 00 − (t + 1)y 0 + y = t2 e3t
t > 0.
Solution:
There are two ways to do this problem. The first method, which works even if
you did not solve part (a), is to observe that while the equation does not have
constant coefficients, the method of undetermined coefficients works as the right
hand side of the equation is a polynomial times and exponential, and the coefficients of the equation are also polynomials. Guess that y = (At2 + Bt + C)e3t .
1
. The general solution is
Plugging in, we find that A = 0, B = 61 and C = − 12
then
1
t
y(t) = c1 (1 + t) + c2 et + e3t ( − ).
6 12
The second method is to use the method of variation of parameters. Let y1 =
(1 + t), y2 = et . We guess Yp (t) = u1 y1 + u2 y2 . Plug in, and set
u01 y1 + u02 y2 = 0.
Compute Yp00 and plug into the equation to obtain
t(u01 y10 + y20 u02 ) = t2 e3t .
Solving we find u01 = −e3t and u02 = (1 + t)e2t . We integrate to find u1 =
−1 3t
e . The integral for u2 is evaluated using integration by parts. We find
3
u2 = ( 2t + 14 )e2t . Combining everything, we obtain the same Yp as above.
4. The population of a protected species of wild salmon in the Pacific Northwest
is modeled by the logistic equation
dP
= P (1 − P ).
dt
(1)
Suppose that, on June 17, 2014 at 5 pm EST, the Department of Fisheries
and Oceans authorizes fishermen to remove K salmon per unit time, as well
as increasing fisheries on the smaller fish the salmon eat. As a result of this
policy change, the new model for the population P is given by
dP
P
= 3P (1 − ) − K,
dt
3
9
K< .
4
(2)
(a) (5 points) Draw the phase line for this new model, and find the equilibria.
Classify the equilibria as stable or unstable.
Solution:
The
graph of dP/dt vs. P is an upside down parabola with zeros at P± =
√
3± 9−4K
. Since K < 94 , this means there are two distinct roots. Then P− is
2
unstable, P+ is stable.
(b) (5 points) Suppose that at 4:59 pm EST on June 17, 2014, just before the
new fisheries act is signed into law, the population P is stable. What is the
population at 4:59pm EST? For what value of K will the new law cause the
population of wild salmon to become extinct? (Remember: before the act is
signed, the population is determined by the model in equation (1). After the
act is signed, the population is determined by the model in equation (2)).
Solution:
The population at the time of signing is stable. Since the population before
the law was signed was given by dP/dt = √
P (1 − P ), we know that P = 1. The
3− 9−4K
. Do some algebra so find this
population will go extinct if 1 < P− =
2
implies that K > 2.
(c) (5 points) Can you suggest some changes to the law which will ensure that
the population of wild salmon never becomes extinct? Provide some evidence
for your suggestions.
Solution:
A sample solution I was hoping for was the following. Rather than remove K
fish, the government could have legislated that the number of salmon allowed
to be fished was a fixed percentage of the total population. This changes the
model to
P
dP/dt = 3P (1 − ) − αP,
3
for some α < 1. This means that the fisherman are allowed to fish α percent of
the total population. In this new model, no matter what the initial population
is (as long as P > 0), the population increases towards the stable equilibrium
at P = (3 − α).
5. (a) (10 points) Find a fundamental set of power series solutions to the equation
(1 + x2 )y 00 − 4xy 0 + 6y = 0,
about the point x0 = 0. Find the first three terms in the expansion, and provide a recursion relation to determine all the coefficients.
Solution: P
n
Guess y = ∞
n=0 an x . Plug this into the equation and reindex to obtain the
recursion relation
(m − 2)(m − 3)
an .
am+2 = −
(m + 2)(m + 1)
Note that am = 0 for all m ≥ 4. The solutions are thus y1 = 1 − 3x2 and
y2 = x − 31 x3 .
(b) (5 points) Consider the equation
2x2 y 00 + 9x cos(xex )ex
2014
y 0 + 6 cos(sin(ex − 1))y = 0.
Suppose that y is any solution of this equation on the set x > 0. We define
A := {b ∈ R : lim+ xb |y(x)| < +∞}.
x→0
Find the set A. (Don’t panic! This problem is easier than it looks.)
Solution: Note that x = 0 is a regular singular point of the differential equation. The first thing to do is compute the indicial equation
2r(r − 1) + 9r + 6 = 0.
This has roots r1 = −3/2 and r2 = −2. Since r1 − r2 = 1/2 is not an integer,
we can find linearly independent solutions
y1 = xr1 p1 (x),
y2 = xr2 p2 (x),
where p1 (x), and p2 (x) are power series, convergent in a neighbourhood of
x = 0, with p1 (0) 6= 0 6= p2 (0). From this, we see
lim xb y(x) = lim+ c1 xb+r1 p1 (x) + c2 xb+r2 p2 (x),
x→0+
x→0
where c1 , c2 are arbitrary coefficients. Since p1 , p2 are continuous, and not zero
at x = 0, this limit is finite whenever b + r1 ≥ 0 and b + r2 ≥ 0. Plugging in
the values of r1 and r2 we see that
A = [2, ∞).