Re: is this solvable?

On Fri, 24 Apr 2009 johnboy98105@xxxxxxxxx wrote:

Consider the following system of ODEs
(first order followed by second order)

{
a' = k1 * b - d
a'' = k1 * b'
}

With following initial conditions and parameters.

a'[0] == 0,
a[0] ==1,
b[0] == 0,

k1 = 1,
r1 = 1
d = 1

Thus in addition to a'(0) = 0,
a'(0) = k1.b(0) - d = -1.

Is that a well posed problem? Is there a solution to that? if so, how
do I approach?

Besides being inconsistent, it's got a heavy computer accent.
In addition what's r1 have to do with it? Here's a math version.

y'(x) = u(x) - 1
y"(x) = u'(x)

y'(0) = 0; y(0) = 1; u(0) = 0

As the second equation is the direct result of the first (assuming
that y' and u are differentiable), it may be discarded.

Do I understand you correctly? You want to solve

y'(x) = u(x) - 1

y'(0) = 0; y(0) = 1; u(0) = 0 ?

Anyway you've got a problem in that u(0) = 0 makes y'(0) = -1.
.

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