Section 2.2: Small changes can result in large changes in the solution. Here are the solutions

to problems 33, 34 and 35 (You'll insert the command for Problem 36)

#Problem 33: (Note: You can insert comments using the # sign)

F1:=int(1/(y-1)^2,y)=x+C;

#Problem 33, solve for C:

solve( subs({x=0,y=1},F1), C);

#Oops! y=1 was an equilibrium solution!

F1:=1;

#Problem 34: I copied off the integral above, and repeated the steps

F2:=int(1/(y-1)^2,y)=x+C;

solve( subs({x=0,y=101/100},F2), C);

solve(F2,y);

F2:=(-1+x-100)/(x-100);

#Space for Problem 35

F3:=int(1/((y-1)^2+0.01),y)=x+C;

solve( subs({x=0,y=1},F3), C);

solve( subs(C=0,F3), y);

F3:=1+0.1*tan(0.1*x);

#Space for Problem 36

#Plot the functions together, close to the point (0,1)

plot([F1,F2,F3],x=-12..12,y=0..2,color=[red,black,blue]);

Section 2.2: Separation of Variables with Level Curves

Here is the solution to Problem 48: (1) Solve y'=(8x+5)/(3y^2+1) (2) Plot the level curves (3) Plot the solutions

corresponding to initial conditions: (0,-1), (0,2), (-1, 4), (-1, -3)

Rather than using Maple's built-in solver, we'll solve this manually using separation of variables. Note

that:

(3y^2+1)dy = 8x+5 dx

Eqn1:=int(3*y^2+1,y)=int(8*x+5,x)+C;

To plot level curves, convert this to the form: f(x,y)=C, so that f(x,y)=y^3+y-4x^2+5x

f:=y^3+y-4*x^2+5*x;

with(plots):

contourplot(f,x=-3..3,y=-3..3);

We can plot the individual solution curves by using "implicitplot" after solving for C. For example, if the initial

condition was (0,-1):

C1:=subs({x=0,y=-1},f);

implicitplot(f=C1,x=-3..3,y=-3..3);

Equivalently, we can define which contours to plot. The first initial condition corresponds to the level curve at -2. We can define the others as well:

C2:=subs({x=0,y=2},f); C3:=subs({x=-1,y=4},f); C4:=subs({x=-1,y=-3},f);

contourplot(f,x=-5..5,y=-5..5,grid=[25,25],contours=[C1,C2,C3,C4]);