Practice Test #1 Mat 242
1.1
(1) Draw a direction field for the given D.E. and determine the behavior of the solution y as t approaches infinity:
y' = 3 - 2y.
(2) What appears to be the equilibrium solution of the D.E. with the given direction field:
1.2
(3) Solve the D.E. (use separation) for the following predator-prey model:
dp/dt = .4p - 800, p(0) = 400.
(4) Radium-226 has a half-life of 1620 years. Find the time period during which a given amount of this material is reduced by one-quarter. (Hint: it decays exponentially).
1.3
(5) Give the order of the D.E., and tell whether or not it is Linear.
(a) d3y - 2 dy = 3 + y
dt3 dt
(b) y'' - y' cos(y - t) = 7/y
(6) If ert is a solution of the D.E. y'' + y' - 6y =0, determine the value of r.
2.1
(7) Find the solution of the D.E.: y' + 2y = te-2t , y(1) = 0.
(8) Find the solution of the D.E.: ty' + (t + 1)y = 1, y(ln 2) = 1.
2.2
Solve the (separable) DE's:
(9) y' = (3x2 - ex) / (2y - 4), y(0) = 3.
(10) Find the solution of the, D.E., and determine the interval for t where the solution is valid:
y' = 3x2/ (2y - 5) y(1) = 6.
2.3
(11) At time t = 0, a tank contains Q0 lb. of salt dissolved in 120 gallons of water. Assume that water containing 1/3 of salt/gal. is entering the tank at a rate of 2 gal/min. and is exiting at the same rate. Set up the initial value problem that models this process, with Q(t) equal to the amount of salt in the tank at any given time t, and find the limiting amount Ql that is present after a long time.
(12)
2.4
Using theorem 2.4.1, determine an interval on which the given DE is certain to have a solution that exists:
(13) (9 - t2) y' + 2ty = 3t2, y(-2) = 1.
(14) (ln t)y' + y = cot (t) y(2) = 3.
2.5
For the given D.E., graph dy/dt = f(y) against y, find all critical points, and determine whether each is asymptotically stable or unstable.
(15) dy/dt = y(y2 -1)
(16) dy/dt = e2y - 1
(17) population problem
2.6
Determine whether the equation is exact, and if it is, find the implicit solution to the equation.
(18) (2xy2 + 2y) + (2x2y + 2x)y' = 0.
(19) x2y3 + x(1 + y2) y' = 0.
Solutions:
(1)
(2) The equilibrium solution is represented by the straight line in the middle, at y = - .5. Notice that the arrows point away from it - this indicates divergence as t approaches infinity.
(3)dp/dt = .4p - 800 is seperable - divide both sides by .4p - 800 and multiply by dt:
dp/(.4p - 800) = dt , then integrate both sides (the right-hand one involves u-substitution, with u = .4p - 800, so that du = .4 dp , or du/.4 = dp, or 2.5dp = du, and we have the integral
2.5∫du/u = 2.5 ln (.4p - 800)) , and we have:
2.5 ln (.4p - 800) = t + C1,
ln (.4p - 800) = (t + C1)/2.5
.4p - 800 = et/2.5 + C
.4p = et/2.5 + C + 800
p = (et/2.5ec+ 800)/.4
p = (Cet/2.5 + 800)/.4, and applying the IC, and if p(0) = 400, we have
400 = (C + 800)/.4,
1000 = C + 800, and
200 = C,
(4) Ok, because this is an exponential decay model, it must be the case that the solution is of the form:
y = Cert , with y(0) = y0 = C our initial amount of radium. If the half-life is 1620 years, that means it takes that long for their to be half of the original amount of radium left, so we'd have:
.5C = Cer1620, and using this equation , by cancelling the C's and taking ln of both sides, we can find the value of r:
ln (.5) = r * 1620
ln (.5) /1620 = r, so
r = -.0004279, and the completed model would be
y = Ce-.0004279t . The question asks when the radium will be reduced by one-quarter. At that time 3/4 will remain, so we'd have
.75C = Ce-.0004279t, the C's would cancel, and we'd get
ln (.75)/-.0004279) = t, or t is about 672.311 years.
(5) (a) The order is 3 (highest-derivative), and the DE is linear (no y's in sin, cos, ln , or e, no y's in denominator, no y to a higher power or terms like yy').
(b) We have a y within sin, and also in a denominator, so the DE is non-linear. It has order 2.
(6) If y = ert , then we would have y' = rert, and y'' = r2ert , and so the equation would become:
r2ert + rert - 6ert = 0, and factoring out ert, we have:
ert(r2 + r - 6) = 0, and setting the polynomial factor = 0, we get the solutions r = -3 and r = 2.
(7) y' + 2y = te-2t , y(1) = 0. Here we have standard form for first-order linear DE's with variable coefficients, with p(t) = 2, and g(t) = te-2t, and the integrating factor is u(t) = e∫2 dt = e2t, and the solution is:
y = (∫e2t (te-2t ) dt + C)/ e2t =( ∫t dt + C)/e2t = (t2e-2t)/2 + Ce-2t, and with the IC y(1) = 0:
12e-2(1) + Ce-2(1) = 0, and so C = -1, and the particular solution is
(t2e-2t)/2 - e-2t.
(8) ty' + (t + 1)y = 1, y(ln 2) = 1. First, we divide by t to get it in standard form:
y' + ((t + 1)/t)y = 1/t, or
y' + (1 + 1/t)y = 1/t, and the integrating factor is u(t) = e∫1 + 1/t dt = et + ln t = et elnt = tet. The solution would be
y = (∫ tet(1/t) dt + C )/tet = (∫et dt + C)/tet = (et + C)/tet = 1/t + C/(tet), and now we apply the IC :
1/(ln 2) + C/(ln (2) e ln 2) = 1, and C = 2 ln(2) - 1.
(9)dy/dx = (3x2 - ex) / (2y - 4), y(0) = 3. Separate by moving the dx to the right, and the (2y - 4) to the left by dividing, and you get
dy/(2y - 4) = (3x2 - ex)dx , and now integrate both sides to get :
(1/2)ln(2y - 4) = x3 + ex + C, and then apply y(0) = 3 to get:
.5 ln (2) = 1 + C, etc.
(10) y' = 3x2/ (2y - 5) y(1) = 6, separate again and we get :
dy/(2y - 5) = 3x2dx , and then integrate both sides for :
.5 ln (2y - 5) = x3 + C, and with y(1) = 6, we get
.5 ln (7) = 1 + C, so that C = -.027
.5 ln (2y - 5) = x3 - .027, now solve it for y:
ln (2y - 5) = 2x3 - .054, and raise e to both sides
2y - 5 = exp(2x3 - .054), and
y = (exp(2x3 - .054) + 5)/2, which is valid for all x.
(11) so change in salt level = rate in - rate out, or
Q' = (1/3)2 - 2(Q/120), and Q(0) = Q0 (Q/120 is the percentage of water in the tank). Put this in the form:
Q' + (1/60)Q = 2/3, and the integrating factor is e∫(1/60)dt =e (1/60)t, and so we have the solution:
Q = (∫e (1/60)tdt + C)/ e (1/60)t = ( 60e (1/60)t + C)/ e (1/60)t , so finally
Q = 60 + C e -(1/60)t , and now use the IC Q(0) = Q0 to get
Q0 = 60 + C e -(1/60)0, and so C = Q0 - 60,
Q = 60 + (Q0 - 60) e -(1/60)t, and if you take the limit as t approaches infinity , it appears this goes to Q = 60 lbs.
(13) (9 - t2) y' + 2ty = 3t2, y(-2) = 1. Get this in standard form by dividing through by (9 - x2), and
y' + (2t/(9 - t2) ) y = 3t2/(9 - t2) , and theorem 2.4.1 guarantees a solution where
(2t/(9 - t2) ) and 3t2/(9 - t2) are continuous, i.e. where t is not equal to 3 or -3, or t < -3, -3 < t < 3, or t > 3. We have to match one of these intervals with the t in the initial condition y(-2) = 1, i.e.
t = -2, and that would be the interval t < -3, so that is where the theorem gurantees a solution.
(14) (ln t)y' + y = cot (t), again get standard form, this time by dividing through by ln t:
y' + y/(ln t) = (cot (t) / ln (t) ) , and find out where the coefficient functions 1/ln t and cot (t) are continuous, i.e. {t > 0, t not equal to 1} for 1/ln (t), and (0, n*pi) for cot (t) , so we have to find an interval matching these restrictions whcih contains our t from the IC y(2) = 3, t = 2. So one of these intervals would be from (1,pi).
(15) (I have to figure out a way to graph it here, but just enter Y1 = x(x2 - 1) on TI-83)
the CP's are at y(y2 - 1) = 0 , or y = 0, y = 1, y = -1. The curve rises through the CP's -1 and 1, and falls through 0, so y = 0 is asymptotically stable, the other two are unstable.
(16) Ok, set e2y - 1 = 0, and you get e2y = 1, take ln of both sides and 2t = ln (1) , or y = 0 is the only CP - now graph on the TI-8x Y1 = e2x - 1, and you'll see that the curve rises through y = 0, so it is not a stable equilibrium point.
(18) (2xy2 + 2y) + (2x2y + 2x)y' = 0, so M(x,y) = 2xy2 + 2y, and N(x,y) =2x2y + 2x, and
Nx = 4xy + 2 = My, and that means that the equation is stable. To find the implicit solution, we take the partial differential equations
fx (x,y) = 2xy2 + 2y, and fy (x,y) = 2x2y = 2x, and integrate both sides of each to get :
f(x,y) = x2y2 + 2xy + h(y) , and f(x,y) = x2y2 + 2xy + h(x), and unioning the terms we have:
f(x,y) = x2y2 + 2xy = c as our implicit solution
(19) (19) x2y3 + x(1 + y2) y' = 0. Since M(x,y) = x2y3 , and N(x,y) = x + xy2 , we have
Nx = 1 + y2 , and My = 3x2y2, which are not equal , and therefore the equation is not exact.