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16b. By taking the derivative of y ln(K/y) it can be shown that dy
the graph of — vs y has a maximum point at y = K/e. Thus dt
— is positive and increasing for 0 < y < K/e and thus y(t) dt
is concave up for that interval. Similarly — is positive
and decreasing for K/e < y < K and thus y(t) is concave down for that interval.
16c. ln(K/y) is very large for small values of y and thus
(ry)ln(K/y) > ry(1 - y/K) for small y. Since ln(K/y) and (1 - y/K) are both strictly decreasing functions of y and since ln(K/y) = (1 - y/K) only for y = K, we may conclude dy
that — = (ry)ln(K/y) is never less than dt
— = ry(1 - y/K). dt
u dy ,,du
17a. If u = ln(y/K) then y = Ke and — = Ke — so that the
D.E. becomes du/dt = -ru.
18a. The D.E. is dV/dt = k - apr . The volume of a cone of
height L and radius r is given by V = pr L/3 where
L = hr/a from symmetry. Solving for r yields the desired solution.
18b.Equilibrium is given by k - apr = 0.
18c. The equilibrium height must be less than h.
20b. Use the results of Problem 14.
20d. Differentiate Y with respect to E. dy
21a. Set — = 0 and solve for y using the quadratic formula. dt
21b. Use the results of Problem 14.
21d. If h > rK/4 there are no critical points (see part a) and dy
— < 0 for all t. dt
1 dx x dn
24a. If z = x/n then dz/dt = — — - — —. Use of
n dt n2 dt
Equations (i) and (ii) then gives the I.V.P. (iii).
24b. Separate variables to get --------------- = - pdt. Using
partial fractions this becomes — + --------- = -pdt.
Integration and solving for z yields the answer.
24c. Find z(20).
26a. Plot dx/dt vs x and observe that x = p and x = q are critical points. Also note that dx/dt > 0 for x < min(p,q) and x > max(p,q) while dx/dt < 0 for x between min(p,q) and max(p,q). Thus x = min(p,q) is an asymptotically stable point while x = max(p,q) is unstable. To solve the D.E., separate variables and use
1 dx dx
partial fractions to obtain ------------- [- - ------] = adt.
q-p q-x p-x
Integration and solving for x yields the solution.
26b. x = p is a semistable critical point and since — > 0,
x(t) is an increasing function. Thus for x(0) = 0, x(t) approaches p as t ^ •. To solve the D.E., separate variables and integrate.
Section 2.6, Page 95
2 2 2 3. M(x,y) = 3x -2xy+2 and N(x,y) = 6y -x +3, so My = -2x = Nx
and thus the D.E. is exact. Integrating M(x,y) with
2 . 6
respect to x we get y(x,y) = x - xy + 2x + H(y).
Taking the partial derivative of this with respect to y
and setting it equal to N(x,y) yields -x2+h'(y) =
6y2-x2+3, so that h'(y) = 6y2 + 3 and h(y) = 2y3 + 3y. Substitute this h(y) into y(x,y) and recall that the equation which defines y(x) implicitly is y(x,y) = c.
3 2 3
Thus x - x y + 2x + 2y + 3y = c is the equation that yields the solution.
5. Writing the equation in the form M(x,y)dx + N(x,y)dy = 0 gives M(x,y) = ax + by and N(x,y) = bx + cy. Now My = b = Nx and the equation is exact. Integrating
M(x,y) with respect to x yields y(x,y) = (a/2)x + bxy +
h(y). Differentiating y with respect to y (x constant) and setting yy(x,y) = N(x,y) we find that h'(y) = cy and
thus h(y) = (c/2)y . Hence the solution is given by
22 (a/2)x + bxy + (c/2)y = k.
7. My(x,y) = excosy - 2sinx = Nx(x,y) and thus the D.E. is
exact. Integrating M(x,y) with respect to x gives
y(x,y) = exsiny + 2ycosx + h(y). Finding yy(x,y) from
this and setting that equal to N(x,y) yields h'(y) = 0 and thus h(y) is a constant. Hence an implicit solution
of the D.E. is exsiny + 2ycosx = c. The solution y = 0
is also valid since it satisfies the D.E. for all x.
9. If you try to find y(x,y) by integrating M(x,y) with
respect to x you must integrate by parts. Instead find y(x,y) by integrating N(x,y) with respect to y to obtain y(x,y) = exycos2x - 3y + g(x). Now find g(x) by differentiating y(x,y) with respect to x and set that