I am trying DSolve in Mathematica for an ODE. Here is the command I am trying and the result I get

DSolve[{x'[t] + y[t] x[t] == 0, x[0] == x0}, x[t], t]

$$ \left\{\left\{x(t)\to \text{x0} \exp \left(\int_1^t -y(K[1]) \, dK[1]-\int_1^0 -y(K[1]) \, dK[1]\right)\right\}\right\} $$

Why does Mathematica separate the integration between 0 and 1 and the rest? The answer as one would expect is

$$ x(t) = x0 \exp\left(-\int_0^t y(t') dt' \right) $$

How to get Mathematica to give the above result from DSolve?

Without the initial condition in the list of equations, I just get the integral from 1 to t.

Thanks for any help or suggestions!

PS: I searched for a solution and couldn't find one. If this was already addressed, please point me to relevant sources.

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    – Michael E2
    Aug 11, 2015 at 16:54

1 Answer 1


DSolve seems to have a general solution to linear first-order ODEs that uses an integration that starts at 1. It then solves the boundary condition for the constant of integration. In this case, the constant involves the integral from 1 to the initial value for t, namely 0.

Block[{DSolve`print = Print},
 DSolve[{x'[t] + y[t] x[t] == 0, x[0] == x0}, x[t], t]

Mathematica graphics

(*  {{x[t] -> E^(-Integrate[-y[K[1]], {K[1], 1, 0}] + 
                  Integrate[-y[K[1]], {K[1], 1, t}])*x0}}   *)

(Mathematica does not have built-in methods for recombining integrals and sums. There are questions about doing such things on this site.)


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