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I have three variables $X, Y, Z$ whose functional form is the following (I also have a function for $Z$ which is somewhat complicated but is not essential for my question here; so it is omitted.):

$X=a+bY$

$\dot{Y}=c+dZ$

where $a,b,c,d$ are abstract parameters, and the overdot refers to time derivative of the variable, i.e. $\frac{dY}{dt}$.

Setting aside an obvious fact that this system is reduced to $\dot{X}=\dot{Z}$, I would like to know the Mathematica code for this system. For the first equation, I have

X[Y_] := a + b*Y

But I am stuck with the second one. Do I have to make the variables a function of time t? But in that case, doesn't the system become discrete, not continuous?

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Yes, make X, Y, and Z functions of t. And use Equal instead of Set or SetDelayed.

Example with Z[t] = Exp[t]:

Block[{Z = Exp},
 DSolve[{X[t] == a + b Y[t], Y'[t] == c + d Z[t]}, {X[t], Y[t]}, t]
 ]
(*  {{X[t] -> a + b (d E^t + c t + C[1]), Y[t] -> d E^t + c t + C[1]}}  *)
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  • $\begingroup$ thank you so much! Question: Why can we not include Z function into the DSolve bracket and instead use Block? $\endgroup$ – ppp Oct 1 '14 at 17:27
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    $\begingroup$ @ppp You're welcome. The C[1] is a constant of integration generated automatically by DSolve; if you add an initial condition for Y, C[1] will be replaced by a value. You can use an arbitrary Z: DSolve[{X[t] == a + b Y[t], Y'[t] == c + d Z[t]}, {X[t], Y[t]}, t] works as is. You said you had a function for Z, so I gave an example with a function for Z. :) $\endgroup$ – Michael E2 Oct 1 '14 at 17:30
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    $\begingroup$ @ppp Oops, I think I misunderstood your Q about Z. Of course you can substitute an expression for Z[t]. Your Q was in terms of Z, so I wanted to give an answer in terms of Z. $\endgroup$ – Michael E2 Oct 1 '14 at 17:31
  • $\begingroup$ I tried DSolve[{X[t] == a + b Y[t], Y'[t] == c + d Z[t], Z[t] == Exp[t]}, {X[t], Y[t], Z[t]}, t] and it says E^t cannot be used as a function. Many thanks :) $\endgroup$ – ppp Oct 1 '14 at 17:43
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    $\begingroup$ @ppp I get {{X[t] -> a + b (d E^t + c t + C[1]), Y[t] -> d E^t + c t + C[1], Z[t] -> E^t}}. (V10.0.1.) $\endgroup$ – Michael E2 Oct 1 '14 at 17:44

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