> 
# A introduction in the use to Maple to solve differential equations.
# Specifically, DEplot is used to get the "feel" for the solutions. Then,
# a sample problem is given which requires one to demonstrate that a 
# particular implicitly defined relation satisfies a differential equation.
# Finally, a  nonlinear 1st order differential equation is given which Maple
# cannot solve using dsolve.
# 
# Written by Doug Hensley
# 
> with(DEtools);

   [DEplot, DEplot1, DEplot2, Dchangevar, PDEplot, dfieldplot, phaseportrait]
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> ?DEplot1
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# 
# Here a solution of # 39 is given: 
> deqn1:=diff(y(x),x)=y(x)*(4-y(x));

                                 d
                      deqn1 := ---- y(x) = y(x) (4 - y(x))
                                dx
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# (go ahead and call on "HELP" for the syntax of DEplot1!)
> DEplot1(deqn1,y(x),x=0..3,y=-1..5);
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# Based on the direction field I got from this, I think I'd conclude that if originally y>0, 
# then y(x) goes to 4 as x goes to infinity, while if y(0)<0 then y(x) goes to -infinity. It 
# doesn't seem to matter what the original x value is...but then it wouldn't: x never 
# appears in the original DE!
# If y=0 originally then y will be zero forever. That's the abstract answer. If this were 
# an applications problem, then we'd have to expect that y gets jostled a bit. Sooner or 
# later, it will escape to 4 or to -infinity.
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# Here's another, harder example. In this problem you try to verify that the equation 
# "eqn" represents an (implicit) solution to the DE "difeq". (Both are given below.) I 
# included some hints, so this is something between a lesson, a sermon, and a 
# homework problem. Please fill in the gaps and satisfy yourself and any reasonable 
# reader that this equation does indeed represent a solution. Note: no claim is made 
# that it's the only solution! There could be other curves, with other starting points. 
# Note also that the solution goes haywire somewhere between 0 and 1. Keep in mind: 
# solutions are FUNCTIONS, defined on INTERVALS.
# 
> with(plots):eqn:=(x=1+(1-y)*ln (1-y));

                        eqn := x = 1 + (1 - y) ln(1 - y)
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> implicitplot(eqn,x=0..1,y=0..1);
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> difeq:=(y(x)-x)*diff(y(x),x)=1-y(x);

                                       /  d      \
                   difeq := (y(x) - x) |---- y(x)| = 1 - y(x)
                                       \ dx      /
> neweqn:=subs(y=y(x),eqn);

                    neweqn := x = 1 + (1 - y(x)) ln(1 - y(x))
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> checkeq:=diff(neweqn,x);

                              /  d      \                /  d      \
             checkeq := 1 = - |---- y(x)| ln(1 - y(x)) - |---- y(x)|
                              \ dx      /                \ dx      /
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# Now some independent pen-and-paper work becomes necessary. What is the 
# connection between difeq, the DE the assignment says matches up with the original 
# equation, and the  actual result "checkeq" of differentiating that equation? 
# 
# It comes to this:
# The DE we KNOW is right is checkeq. Also, "neweq" is given. The DE we WANT 
# is difeq. Why do checkeq and neweq, together, imply difeq?
# The key is in the repeated appearance of the expression ln(1-y(x)). You can solve 
# neweq for ln(1-y(x)) in terms of the other ingredients of neweq. Then, you can plug 
# the result of this calculation in for ln(1-y(x)) in checkeq. With any luck, it should 
# now look a lot like difeq.
# I suppose it would be possible to go back and make it look as though this part of the 
# problem could also have been delegated to Maple. I don't see how offhand. I rather 
# think it's at least as hard to write the code that will cause Maple to confirm the 
# answer as it is to just finish up by hand.
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> difeq2:=diff(y(x),x)=solve(difeq,diff(y(x),x));

                                   d           - 1 + y(x)
                       difeq2 := ---- y(x) = - ----------
                                  dx            y(x) - x
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> expn:=rhs(difeq2);

                                        - 1 + y(x)
                              expn := - ----------
                                         y(x) - x
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> dsolve({diff(y(x),x)=expn,y(1)=0},y(x));
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# Well, it seems as though Maple can't be used as an oracle. There is no output from 
# this command, which means that the program hasn't anything to report.
>