I do not understand why the following code does not solve the equation:


eqn = y[t] == \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)]\(Exp[
      a \((t - s)\)] y[s] \[DifferentialD]s\)\);
sol = DSolve[eqn, y[t], t, y[0] = 1]

Can someone help me please? Thanks in advance.

I tried to use yours idea on the real case, obtaining the following, which is still an integral-differential equation:


  • $\begingroup$ DSolve[]'s support for integral equations is still somewhat limited, so don't be surprised if some things don't work yet $\endgroup$ – Mariusz Iwaniuk Oct 26 '19 at 15:09
  • $\begingroup$ Are there some other functions which could be useful? $\endgroup$ – Umberto Tomasini Oct 26 '19 at 15:22
  • $\begingroup$ Use LaplaceTransform for equation. $\endgroup$ – Mariusz Iwaniuk Oct 26 '19 at 15:28

You can sometimes remove the integral part by differentiating the expression with respect to $t$. For example,

D[y[t] == Integrate[Exp[a*(t - s)]*y[s], {s, 0, t}], t]
(*y'[t] == Integrate[a*E^(a*(-s + t))*y[s], {s, 0, t}] + y[t] *)

We see that the first term on the right hand side is simply $ay(t)$. So, now you can solve the differential equation



DSolve[{y'[t] == (a + 1) y[t], y[0] == 1}, y[t], t]
(* {{y[t] -> E^(t + a t)}} *)
| improve this answer | |
  • $\begingroup$ I agree with you, the fact is that I was looking for solving a more difficult integral equation, which related differential equation still presents integrals which I cannot neglect. $\endgroup$ – Umberto Tomasini Oct 26 '19 at 18:15
  • $\begingroup$ To clarify, I did not neglect the integral. If your kernel function is separable (i.e., $k(t,s)=f(t)g(s)$), then the following relation should hold upon differentiation y'[t] == y[t] (f[t] g[t] + f'[t]/f[t]). No integrals are necessary. If it is not separable, then, as Mariusz stated, your options are probably limited, but you would need to provide an example to be sure. $\endgroup$ – Tim Laska Oct 27 '19 at 0:21
  • $\begingroup$ Yes, you are right, I get your idea. I tried to apply it on the real case, obtaining the differential-integral equation above $\endgroup$ – Umberto Tomasini Oct 27 '19 at 14:42
  • $\begingroup$ You should provide the initial definition for $y(t)$ and $K(t-s)$. It will be easier for people to help if you also provide the Mathematica code. $\endgroup$ – Tim Laska Oct 28 '19 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.