306D Math Lab #3 Problem #1: y t c is an implicit solution of the logistic 1y 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) . Be sure to include the constant solutions of y' y(1 y) in your phase portrait. Text #2= Differential Equations with Maple by Coombes, Hunt, Lipsman, Osborn, Stuck C.O. Bloom C.O. Bloom