Re: --Ping Dave, changed my mind - I am still confused

On Feb 9, 7:19 pm, "The poster formerly known as Colleyville Alan"
<nos...@xxxxxxxxxx> wrote:
...
But, during the process of integration, I get ln|y| = -t + c.  When I
exponentiate both sides, I get |y| = e^(-t+c).  Now, according to the
definition of abs value, if the contents of y are positive, then |y| = y and
if the contents of y are negative, then |y| = -y.  Since the RHS has a
positive e raised to some exponent, this value is positive and therefore y
would be positive.  

This is incorrect. What is the solution of |y| = 1? It is y = +- 1,
because |1| = 1 and |-1| = 1. The fact that the right hand side is
positive does not mean that y is positive. Similarly, |y| = e^(-t+c)
means that y = +- e^(-t+c), because |e^(-t+c)| = e^(-t+c) and |-e^(-t
+c)| = e^(-t+c).

If I had raised a *negative* e to some power, then I
would say |y| = -y, but since this is a positive e on the RHS, I do not see
how that is possible in this case.

This doesn't follow. For example, the equation |y| = -1 has no
solutions.

I realize that the initial conditions require that c = -1 but if c_2 =
e^c_1, then I do not see how c_2 can be = 0 in this case.  So, even though I
replaced e^c_1 with c_2, since c_2 represents e^c_1, it does  not seem to me
that I can simply treat it like an arbitrary constant; it is  a constant,
but one that is +e raised to some power of an arbitrary c_1.

Right. But this is where the +- sign comes in. c_2 is positive, so
+c_2 is positive and -c_2 is negative. Writing c_3 = +-c_2, then c_3
can be positive if you choose the + sign and negative if you choose
the - sign. Thus we can write y = 2 + t + c_3 e^-t, where c_3 is an
arbitrary constant. If c_3 turns out to be negative, as it does in
this case to fit the initial conditions, that simply means that we
used the - sign in +-c_2. Recapping your various c values: c_3 = -1,
c_2 = 1, and c_1 = 0, because -1 = - 1 = -e^0.

Dave
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