MA 1270: EXAM 1 Review

Problems similar to the following will be given in the exam.

Selected hints

Modeling problem.

Let X,Y be the two species. Then a possible model is

dX/dt = -k1 X + k2(1-X/N)X Y

dY/dt = -k3 Y + k4 X Y

The first terms in each equation are the death rates. The second terms in each equation are the growth due to the interaction. Note that species X has a growth rate than hits zero when X=N and is negative if X>N due to growth limited crowding.

Separation of variables.

dx/(x^2 + 1) = t dt

Integration yields

atan(x) = (t^2)/2 + C

The initial condition implies that C=pi/4 since atan(1)=pi/4. Thus

x(t) = tan((t^2)/2 + pi/4)

Bifurcation problem.

Recall that a bifurcation occurs if f(x,mu)=0 and df(x,mu)/dx=0. So

df(x,mu)/dx = exp(-x) - x exp(-x) = 0 implies

(1-x) exp(-x) = 0 so

x=1

Thus the critical value of x is 1 and since f(1,mu)=0, this means, mu=1 exp(-1) = 1/e. The diagram is

Nth order ODEs

The characteristic polynomial is x^2(x^2+4)^2. x=0 is a double root and x=2i, -2i are also double roots. So the general solution is

y(t)= A+Bt + (C+Dt)sin(2t)+(E+Ft)cos(2t)

The Routh Hurwitz criterion for third order equations implies that

p^2>0, 1-p > 0, p^2 -(1-p) > 0
Solving the quadratic inequality (finding the roots of the quadratic) means that

(-1/2 + sqrt(5)/2) < p < 1

Linear 1st order ODEs

Recall that the general solution is

x(t) = exp(A(t)) [ C + int{ exp(-A(t)) b(t) dt}]

A(t) = int{a(t) dt}.

Thus, for the present problem, A(t) = 2 ln(t) = ln(t^2), so

x(t) = t^2 [ C + int{ (1/t^2) t^2 dt}] = C t^2 + t^3

The initial condition implies that C=3 so x(t)=3t^2+t^3.