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Midterm ExamMath S3027 - T. Collins

Solve all 5 problems.

1. (10 points) Find the general solution of the initial value problem

t3y + 4 t2y = et + 1 , t > 0.

Solution:Use the method of integrating factors. Let (t) be the integrating factor, thenwe nd (t) = e4ln( t ) = t4. Integrate the equation to obtain

y(t) = Ct4 + 12

t2 (1 + t)t4et .

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2. Consider the initial value problem:

y y 12y = 0, y(0) = a > 0, y (0) = 1.

(a) (10 points) Solve the initial value problem.

Solution:The characteristic equation is r2 r 12 = 0, or (r 4)(r + 3) = 0. As aresult, we nd y1 = e4t and y2 = e3t . Then y = C 1y1 + C 2y2. Plugging in theinitial condition we obtain C 1 + C 2 = a and 4C 1 3C 2 = 0. Solve this systemto obtain C 1 = 3a 17 , and C 2 =

4a +17 .

(b) (5 points) Find the critical value of a that separates solutions thatapproach + as t + from solutions that approach as t + .

Solution:Since e3t is bounded as t + , and e4t as t + , we need to deter-mine the sign of C 1. If C 1 > 0, then y + , and if C 1 < 0 then y + .From this we see that a = 13 is the critical value.

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3. (a) (10 points) Consider the differential equation

t2y t(t + 2) y + ( t + 2) y = 0 t > 0.

Given that y1(t) = t is a solution, nd a second, linearly independent solution.

Solution:Use the method of reduction of order. We guess y2(t) = u(t)t. Plug this intothe equation to nd that y2(t) is a solution provided

t3u t3u = 0.

Solve this equation by integration to obtain u = et . As a result we obtainy2(t) = te t .

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(b) (10 points) Find the general solution of the inhomogeneous equation

t2y t(t + 2) y + ( t + 2) y = 2t3 t > 0.

Solution:There are two ways to do this problem. The rst method is by cleverly observ-ing that, while the equation does not have constant coefficients, the method of undetermined coefficients will still work, as the right hand side of the equationis a polynomial, as are the coefficients of the equation. Since the right handside is a cubic polynomial, and the coefficient of y is order two, we are leadto guess that Y p(t) = At 2 + Bt + C . Plugging in, we nd that B = C = 0 andA = 2. The general solution is then y(t) = c1t + c2te t 2t2.

Alternatively, you can use the method of variation of parameters. We guessY p(t) = u1t + u2te t . Compute Y p = u1 + tu1 + u2(et + te t ) + u2te t . We set allterms containing a derivative of u1 or u2 to be zero;

tu 1 + tet u2 = 0.

Compute Y p and plug into the equation to obtain

t2(u1 + ( et + tet )u2) = 2 t

3.

Solve these two equations for u1, u2 and integrate to nd

u1 = 2t, u2 = 2et .

As a result, we getY p(t) = 2t2 2t.

This gives the same general solution as above.

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4. The population of a protected species of wild salmon in the Pacic Northwestis modeled by the logistic equation

dP dt = P (1 P ).

Suppose that, on June 18, 2013 at 5 pm EST, the Department of Fisheriesand Oceans authorizes shermen to remove K salmon per unit time, as wellas increasing sheries on the smaller sh the salmon eat. As a result of thispolicy change, the new model for the population P is given by

dP dt

= 3P (1 P 3

) K, K < 94

.

(a) (5 points) Draw the phase line for this new model, and nd 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 94K 2 . Since K 2.

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(c) (5 points) Can you suggest some changes to the law which will ensure thatthe population of wild salmon never becomes extinct? Provide some evidencefor your suggestions.(You dont need to ll this page.)

Solution:I basically took anything here, as long as it was motivated in some way by themodel. A sample solution I was hoping for was the following.Rather than remove K sh, the government could have legislated that thenumber of salmon allowed to be shed was a xed percentage of the totalpopulation. This changes the model to

dP/dt = 3P (1 P 3

) P,

for some < 1. This means that the sherman are allowed to sh percent of the total population. In this new model, no matter what the initial populationis (as long as P > 0), the population increases towards the stable equilibriumat P = (3 ).

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5. (a) (10 points) Find a fundamental set of power series solutions to the equation

(2 + x2)y xy + 4 y = 0,

about the point x0 = 0. Find the rst three terms in the expansion, and pro-vide a recursion relation to determine all the coefficients.

Solution:Guess y = n =0 an xn . Plug this into the equation and reindex to obtain therecursion relation

an +2 = (n2 2n + 4)2(n + 2)( n + 1)

an .

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(b) (5 points) Consider the equation

2x2y + 9x cos(xex )ex2013

y + 6 cos(sin( ex 1))y = 0.

Suppose that y is any solution of this equation on the set x > 0. We dene

A := {b R : limx0+

xb|y(x)| < + } .

Find the set A. (Dont panic! This problem is easier than it looks.)

Solution: Note that x = 0 is a regular singular point of the differential equa-tion. The rst thing to do is compute the indicial equation

2r (r 1) + 9 r + 6 = 0 .

This has roots r1 = 3/ 2 and r2 = 2. Since r1 r 2 = 1/ 2 is not an integer,we can nd linearly independent solutions

y1 = xr 1 p1(x), y2 = xr 2 p2(x),

where p1(x), and p2(x) are power series, convergent in a neighbourhood of x = 0, with p1(0) = 0 = p2(0). From this, we see

limx0+

xby(x) = limx0+

c1xb+ r 1 p1(x) + c2xb+ r 2 p2(x),

where c1, c2 are arbitrary coefficients. Since p1, p2 are continuous, and not zeroat x = 0, this limit is nite whenever b + r1 0 and b + r2 0. Plugging inthe values of r 1 and r2 we see that

A = [2, ).