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I have two differential equations:

R * (dVi/dt) + (1/C) * Vi + P0 = P1 , 0<=t<=t[i]
R * (dVe/dt) + (1/C) * Ve + P0 = 0 , t[i]<=t<=t[tot]

With some help I solved the first one

ClearAll[Vi, t, C0, p0, p1, R0];
ode = R0*Vi'[t] + (1/C0)*Vi[t] + p0 == p1
ic = Vi[0] == 0
sol = DSolve[{ode, ic}, Vi[t], t]

But the second one has as initial condition 

Ve(t[i])=Vi(t[i])

I tries to solved it like this knowing that Ve(t[tot])=0 but it doesn't work

ode2 = R0*Ve'[t] + (1/C0)*Ve[t] + p0 == 0
ic2 = Ve[t] == Vi[t]
sol2 = NDSolve[{ode2, ic2}, Ve[t], {t, sol, 0}]
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  • $\begingroup$ Initial conditions can't really be written as ic2 = Ve[t] == Vi[t]. You need to give it a specific value at some specific time. may be ic2 = Ve[0] == Vi[0] or something like this. I also do not understand Ve(t[i])=Vi(t[i]) what is i here? $\endgroup$
    – Nasser
    Commented Nov 19, 2017 at 0:08
  • $\begingroup$ @Nasser i didn't write it right, I shouldn't write it as t[i] just ti which is a moment in time . I have two intervals, [0,ti] and [ti,ttot] $\endgroup$
    – Darius Ionut
    Commented Nov 19, 2017 at 0:33
  • $\begingroup$ Well, when you solve an ODE by hand, the constant of integration is found from initial conditions. So you need to give some specific time value. Something like Ve[5]==someValue, Otherwise, I am not sure how giving an interval is going to work. I never seen an ODE solved using interval for initial conditions. $\endgroup$
    – Nasser
    Commented Nov 19, 2017 at 0:36
  • $\begingroup$ @Nasser cit-evolution.weebly.com/uploads/6/8/9/8/6898684/… the problem is at page 99 in the pdf, if you have some time to look at it, maybe I don't understant the text enought but I relly don't know how to write it $\endgroup$
    – Darius Ionut
    Commented Nov 19, 2017 at 0:45
  • $\begingroup$ @DariusIonut two struggling things: 1.- there is no problem like this one you are asking for help in p. 99 of the book you mention. 2.- how is possible that the second ODE solution will evolve from a time t[i] to 0 ({t, sol ,0})? $\endgroup$
    – José Antonio Díaz Navas
    Commented Nov 19, 2017 at 17:46

1 Answer 1

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You have to use the output of the first ODE at time ti as the initial condition for the second ODE.

ODE #1

DSolveValue[
 {
  p0 + vi[t]/c0 + r0 Derivative[1][vi][t] == p1,
  vi[0] == 0
  },
 vi[t],
 t]

(* -c0 E^(-(t/(c0 r0))) (-1 + E^(t/(c0 r0))) (p0 - p1) *)

Now we use the answer to define vi[t].

vi[t_] := -c0 E^(-(t/(c0 r0))) (-1 + E^(t/(c0 r0))) (p0 - p1)

We evaluate it at time ti to be used as the initial condition for the second ODE.

vi[ti]

(* -c0 E^(-(ti/(c0 r0))) (-1 + E^(ti/(c0 r0))) (p0 - p1) *)

ODE #2

We use the value of vi[ti] as the initial condition in the second ODE.

DSolveValue[
 {
  p0 + ve[t]/c0 + r0 Derivative[1][ve][t] == 0,
  ve[ti] == -c0 E^(-(ti/(c0 r0))) (-1 + E^(ti/(c0 r0))) (p0 - p1)
  },
 ve[t],
 t]
(* -c0 E^(-(t/(c0 r0))) (-p0 + E^(t/(c0 r0)) p0 + p1 - E^(ti/(c0 r0)) p1) *)

and use the answer to define the function ve[t]

ve[t_] := -c0 E^(-(t/(c0 r0))) (-p0 + E^(t/(c0 r0)) p0 + p1 - E^(ti/(c0 r0)) p1)

Now combine the two functions into one (a voltage?).

v[t_] := Piecewise[{
   {vi[t], t <= ti}
   },
  ve[t]
  ]

Set some values for the parameters and plot the result.

r0 = 5;

p0 = 1;

c0 = 1;

p1 = 3;

ti = 1;

Plot[v[t], {t, 0, 2}, PlotStyle -> Black]

Mathematica graphics

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