Math Lab #3
Problem #1:
306B
Spring 2003
y
 t  c is an implicit solution of the logistic
1y
equation (2) y'  y(1  y) . This implicit solution can be derived by the method of separation of
variables (Can you derive it ?). To satisfy the initial condition y(0)  y0 we set
For any constant c the equation (1) F(y)  ln
y0
 (4) F(y)  F(y0 )  t . Equation (4) is the implicit solution of the initial
1  y0
value problem y'  y(1  y) , y(0)  y0 . More generally (5) F(y)  F(y0 )  t  t 0 is an implicit
solution of the initial value problem y'  y(1  y) , y(t 0 )  y0 .
(a) If we use the command "implicitplot" ( in the "plots" package) to sketch F(y)  F(y0 )  t  t 0
for a fixed value of y0 between 0 and 1 ( 0  y0 1 ) and for every value of t0 we generate the phase
portrait of the o.d.e. in the infinite, horizontal strip of the t-y plane between the constant solutions
y  0 and y  1 .
Let y0 .5 and use "implicitplot" to sketch F(y)  F(y0 )  t  t 0 for t0  2, 1,0,1,2 . You
should obtain
Figure 1:
(3) c  F(y0 )  ln
(b) If we use the command "implicitplot" to sketch F(y)  F(y0 )  t  t 0 for a fixed value of y0
greater than 1 (1  y0 ) and for every value of t0 we generate the phase portrait of the o.d.e. in the
infinite, horizontal strip of the t-y plane lying above the constant solutions y  1 .
Let y0  2 and use "implicitplot" to sketch F(y)  F(y0 )  t  t 0 for t0  2, 1,0,1,2 . You
should obtain
Figure 2:
(c) If we use the command "implicitplot" to sketch F(y)  F(y0 )  t  t 0 for a fixed value of y0
less than 0 ( 0  y0 ) and for every value of t0 we generate the phase portrait of the o.d.e. in the
infinite, horizontal strip of the t-y plane lying below the constant solutions y  0 .
Let y0  1 and use "implicitplot" to sketch F(y)  F(y0 )  t  t 0 for t0  2, 1,0,1,2 .
You should obtain
Figure 3
(d) Use the "display " command to combine the above graphs into a complete (global) phase portrait
of y'  y(1  y) .
-----------------------------------------------------------------------------------------------------------------------Math Lab #3
306B
Spring 2001
Problem #2:
(a) Use the Maple command fsolve to locate the equilibrium points of the o.d.e.
y ' = f(y) = sinh(y) sinh(1-y) .
(Recall that sinh(y) = ( exp(y) - exp(-y) )/2 and that cosh(y) = ( exp(y) + exp(-y) )/2.)
(b) Use the Maple command plot to sketch the graph of f(y) = sinh(y) sinh(1 - y) .
(c) Use the Maple command diff to calculate f '(y) and the command fsolve to locate
the point or points where f '(y) = 0.
(d) Use the Maple command evalf to evaluate f '(y) at each equilibrium point found
in (a), and use the results to classify each equilibrium point.
(e) The equilibrium solutions divide the t-y plane into strips.
On the same graph sketch (i) the equilibrium solutions of the
given o.d.e., (ii) one or more solutions of the given o.d.e. that lie
in each strip. Use the Maple command DEplot in the DEtools package.
(f) Use the Maple command DEplot or dfieldplot to sketch the direction field of
the given o.d.e.
(g) Briefly compare the qualitative behavior of the solutions of
the given o.d.e. to the qualitative behavior of the solutions of
y ' = y (1 - y ).
Text #1= Differential Equations by Blanchard, Devaney and Hall
Text #2= Differential Equations with Maple by Coombes, Hunt, Lipsman, Osborn, Stuck
C.O. Bloom
C.O. Bloom