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

equation

ClearAll
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:

eq

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  • $\begingroup$ 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
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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

$$y'(t)=(a+1)y(t)$$

Or

DSolve[{y'[t] == (a + 1) y[t], y[0] == 1}, y[t], t]
(* {{y[t] -> E^(t + a t)}} *)
| improve this answer | |
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  • $\begingroup$ 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

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