Chapter 2, Math 2250 or 2280, EdwardsPenney textbooks
Section 2.1 ==================
Updated 2015
2.18: Solve y'=7y(y13) with y(0)=17
======
Step 1. Identify as variables separable y'=F(x)G(y), with F=7,
G=y(y13).
Step 2. Divide by G(y) and apply quadrature to get 7x+c on the
right and a partial fraction integration problem on the
left:
 y'dx 
int = 7x+c
 y(y13) 
1 A B
Step 3. Partial fractions  =  + 
y(y13) y y  13
Use a sampling method to solve for A=/13, B=1/13.
Step 4. Evaluate the constant c using y(0)=17.
Step 5. Solve for y(x), the explicit solution. Use log and
exponential rules and the trick u = +1 or 1 times u.
Step 6. Answer check. The book answer is correct. The text also has
a phase diagram drawn, which matches what you could do in
two minutes from the phase line diagram method in class.
See textbook section 2.2 for more info.
MAPLE ANSWER CHECK
# y'=7y(y13) with y(0)=17
f:=(x,y)>7*y*(y13);
de:=diff(y(x),x)=f(x,y(x)); # Symbol y(x), not y!
ic:=y(0)=7;
sol:=dsolve([de,ic],y(x));
ans:=rhs(sol);
plot(ans,x=0..0.5,y=0..1);
# answer 91/(7+6*exp(91*x))
2.116: Solve P'=aPbP^2, given P(0)=120 and aP=8, bP^2=6 at t=0.
=======
The rabbit problem. Information in English in the problem statement
implies P'=aP  bP^2, P(0)=120. The terms aP and bP^2, known as
birth and death terms, are given at t=0: aP(0)=8, b(P(0))^2=6.
Insert P(0)=120 to find a=1/15 and b=6/(120)^2. Let M=a/b=carrying
capacity. Then M=160. Use the book formula in section 2.1 for P(t):
M P(0)
P(t)= 
P(0)+(MP(0))exp(a t)
Fill in all the constants and then solve P(T)=0.95 M for T=15
ln(95/15).
M P(0)
P(T)= 
P(0)+(MP(0))exp(a T)
95M M P(0)
 =  Cancel M
100 P(0)+(MP(0))exp(a T)
95 120
 =  Subst P(0)=120,
100 120+(160120)exp(T/15) a=1/15, M=160
To Finish: Crossmultiply and isolate exp(T/15) on the left. Take
the log of both sides, simplify, and find the value of unknown T.
MAPLE ANSWER CHECK
# P'=aPbP^2 with a=1/15, b=6/120^2, P(0)=120
a:=1/15:b:=6/120^2:
f:=(t,P)>a*Pb*P^2;
de:=diff(P(t),t)=f(t,P(t)); # Symbol P(t), not P!
ic:=P(0)=120;
sol:=dsolve([de,ic],P(t));
ans:=rhs(sol);
plot(ans,t=0..100);
# ans := 480/(3+exp(t/15))
Section 2.2 ==================
Updated 2015
2.210: Solve x'= x^2 + 7x  10
=======
Factor the quadratic as (x^27x+10) = (x5)(x2). Substitute
P=x5 or P=x2 to make it into a Verhulst DE. For P=x2 we would
get Verhulst DE
P' = (x  2 3)(x2) = (P3)(P) = (3P)P
Then P'=(abP)P with a=3 and b=1. Plug a,b into the equation
a P_0
P(t) = 
bP_0 + (ab P_0)exp(at)
to find P(t) and then backsubstitute to obtain
x(t) = 2 + P(t) = something like the BOB answer.
The twist: P_0 = x_0 2, which requires even more algebra, finally
matching the book's answer.
MAPLE ANSWER CHECK
# x'= x^2 + 7x  10
f:=(t,x)>x^2+7*x10;
de:=diff(x(t),t)=f(t,x(t)); # Symbol x(t), not x!
sol:=dsolve(de,x(t));
# answer x(t) = (5*exp(3*t)*_C12)/(1+exp(3*t)*_C1)
# Symbol _C1 is an arbitrary constant, call it c. Then
# x(t) = (5*c*exp(3*t)2)/(c*exp(3*t)1)
2.214: Given x' = x(x^2  4)
====== Make a phase line diagram and a phase diagram
Add labels: spout, funnel, node, stable, unstable
Don't use a computer.
Method:
Step 1. Find the equilibria x=0, x=2, x= 2
Step 2. Add them to the phase line diagram
Step 3. Evaluate x' = x(x^2  4) at t=0, x=x_0 where x_0 is
located between equilibria.
Example: for 0 < x_0 < 2, choose x_0 = 1 (a point between)
Then

x' = x(x^2  4)  = 1(1^2  4) = negative
 x = 1
Step 4. Add the PLUS and MINUS signs to the phase line diagram.
Step 5. Draw the phase diagram.
First, draw the equilibria, they are horizontal lines
x = 0, x = 2, x = 2
Second, draw threaded edgetoedge curves, based on the PLUS
and MINUS signs. Draw an increasing threaded curve for PLUS.
Draw a decreasing threaded curve for MINUS.
Step 6. Identify spout, funnel and node from geometry. Then
add STABLE to the funnel and UNSTABLE to the others. Put all
labels onto the phase diagram.
Reminder: Don't use a computer! This problem is a template for
Midterm 1 problem 5.
2.218: Given x' = x^3(x^2  4)
Make a phase line diagram and a phase diagram
Add labels: spout, funnel, node, stable, unstable
Don't use a computer.
Method:
Step 1. Find the equilibria x=0, x=2, x= 2
Step 2. Add them to the phase line diagram
Step 3. Evaluate x' = x^3(x^2  4) at t=0, x=x_0 where x_0 is
located between equilibria.
Example: For 0 < x_0 < 2, choose x_0 = 1 (a point between)
Then

x' = x^3(x^2  4)  = 1(1^2  4) = negative
 x = 1
Step 4. Add the PLUS and MINUS signs to the phase line diagram.
Step 5. Draw the phase diagram.
First, draw the equilibria, they are horizontal lines
x = 0, x = 2, x = 2
Second, draw threaded edgetoedge curves, based on the PLUS
and MINUS signs. Draw an increasing threaded curve for PLUS.
Draw a decreasing threaded curve for MINUS.
Step 6. Identify spout, funnel and node from geometry. Then
add STABLE to the funnel and UNSTABLE to the others. Put all
labels onto the phase diagram.
Reminder: Don't use a computer! This problem is a template for
Midterm 1 problem 5.
Section 2.3 ==================
Updated 2015
2.310:
=======
Prior to chute open, the velocity model is v'=320.15v, v(0)=0. The
distance model is y'=v(t), y(0)=10000. The answers are v(20) approx 202
and y(20) approx 7084.
After chute open, the velocity model is w'=321.5w, w(0)=v(20). The
distance model is x'=w(t), x(0)=y(20). Some maple solving is required,
or a graphing calculator with zoom, to find the answer t approx 326.
The book's answer is 300 min (5 min) plus 47 seconds. This is obtained
by adding 20 sec to the 326 sec found after chute open.
=====maple example=====
eq:= 120*exp(1.5*t)+21*t;
solve(eq=7000,t); # solve an equation for t
# This code must be modified to solve problem 2.310
2.320:
=======
The solution uses book formulas. There is no solving of a DE, because
that has been done already in the textbook, section 2.3.
The model is v'=gr v^2, v(0)=160 where g=32 and r=1/800. This model
is solved on page 101. Do not solve the model; use formulae (13), (14)
page 103 with v0=160, g=32, rho=1/800. The symbol y(t) is the height of
the bolt measured from the ground, so y(0)=0.
The maximum height is obtained for a value of t found by setting the
velocity zero, using equation (13). Because tan(theta)=0 at theta=0,
then the tangent argument in (13) must be zero:
C_1  t sqrt(rho g) = 0.
Solve this equation for t, then make t into a decimal number using
C_1 = arctan(v_0 sqrt(rho/g)).
Use equation (14) with the computed value of t to find the maximum
height. Answer 277.26 ft published in the BOB is correct.
2.322:
=======
The formulas in the textbook are used. No solving of the DE is
required.
The model is v'=g+r v^2 where g=32 and r=0.075 (Downward Motion, page
104). Because the parachutist bails, v(0)=0 and y(0)=10000=y0, where
y(t) is the distance from the ground. Equations (16) and (17) require
v(t)<0, which means y'(t)<0 or y(t) decreasing. Opposite coordinates in
which y(t) increases and measures the distance from the airplane would
require a different model, not found in the textbook. Use formulae
(17), (18) to answer the questions. Do not solve the DE; that is done
in the book. Formula (17) gives 0=y0(1/r)ln(u(t)) where
y0=10000, r=0.075, u(t)=cosh(c2t sqrt(rg))/cosh(c2).
Solve for t to find the flight time. The terminal velocity is given by
(18), the logic from (16), in which tanh(theta) approaches 1 at theta=
infinity.
DEFINITIONS.
cosh(u) is the hyperbolic cosine, defined by cosh(u)=(exp(u)+exp(u))/2
sinh(u) is the hyperbolic sine, defined by sinh(u)=(exp(u)exp(u))/2
tanh(u)=sinh(u)/cosh(u)
arctanh(u) is defined by arctanh(tanh(x))=x
Similarly for arcsinh, arccosh.
maple details =======
u:=t>cosh(c2t*sqrt(rho*g))/cosh(c2);
eq:=10000(1/rho)*ln(u(t));
solve(eq=0,t); # solve an equation for t
maple end ======
=== DEplot notes ===
with(DEtools):
de:=diff(x(t),t)=x(t)*(2x(t)):
var:=x(t):tDomain:=t=0..5:
dots:=[[x(0)=1],[x(0)=1.5],[x(0)=2.2],[x(0)=0.2]]:
wind:=x=2..3:
opts:=linecolor=BLACK,thickness=3,stepsize=0.1:
DEplot(de,var,tDomain,dots,wind,opts);
# This plot can also be carried out from a GUI Tool in maple 12 and 13.
# To find it, start java maple 12 or 13, then goto menu item
# TOOLS ==> ASSISTANTS ==> ODE ANALYZER
# An equivalent command in either a java or nonjava worksheet is
dsolve[interactive]();
# Once started, enter the differential equation as
## diff(x(t),t)=x(t)*(2x(t))
# and the initial condition as
## x(0)=1.5
# Try buttons
## SOLVE SYMBOLICALLY
### Click on SHOW MAPLE COMMANDS
## SOLVE
## PLOT
=== DEplot notes 2 ===
with(DEtools):
de:=diff(x(t),t)=x(t)*(1x(t)):
vars:=x(t):
wind:=t=0..5,x=0.2..3:
ic:=[[x(0)=1.2],[x(0)=1],[x(0)=0],[x(0)=0.5]]:
opts:=stepsize=.2,title=`Verhulst model`,arrows=MEDIUM:
DEplot(de,vars,wind,ic,opts);
# This plot can also be carried out from a GUI Tool in maple 12 and 13.
# To find it, start java maple 12 or 13, then goto menu item
# TOOLS ==> ASSISTANTS ==> ODE ANALYZER
# An equivalent command in either a java or nonjava worksheet is
dsolve[interactive]();
# Once started, enter the differential equation as
## diff(x(t),t)=x(t)*(1x(t))
# and the initial condition as
## x(0)=1.2
# Try buttons
## SOLVE SYMBOLICALLY
### Click on SHOW MAPLE COMMANDS
## SOLVE
## PLOT
Section 2.4 ==================
See numerical links at the course web site.
Euler's method, also Rectangular Rule
Section 2.5 ==================
See numerical links at the course web site.
Improved Euler or Heun's method, also Trapezoidal Rule
Section 2.6 ==================
See numerical links at the course web site.
RungeKutta 4 (RK4), also Simpson's Rule