# Solving an Integro-differential equation numerically

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Freyja on 28 Mar 2015
Answered: Roger Stafford on 9 Jan 2017
Hi, I am interested in writing a code which gives a numerical solution to an integro-differential equation. First off I am very new to integro-differential equations and do not quite understand them so I decided to start simple and would like some help with the first steps. My proposed equation is in the attached picture and the formulas I wish to use are also there though I'm open to suggestions. Even if someone can help me with the first step (just the maths part) where i = 0 I would be very grateful. My goal is to end up with a system of linear algebraic equations which I can then solve with Matlab. Thanks in advance to anyone who takes the time to look at this tricky problem.
Best regards, Freyja Claudio Gelmi on 6 Jan 2017
Edited: Claudio Gelmi on 9 Jan 2017
Take a look at this solver:
"IDSOLVER: A general purpose solver for nth-order integro-differential equations": http://dx.doi.org/10.1016/j.cpc.2013.09.008
Best wishes,
Claudio

Roger Stafford on 9 Jan 2017
I hate to see numerical approximation methods used when there exists a very simple and precise method done by hand. First we designate by K the integral of t*y(t) from 0 to 1, which is unknown as yet. This gives
y(x) = 1 + (K-1/3)*x
Integrating this w.r. to x gives
y(x) = x + (K-1/3)*x^2/2 + C
where C is the unknown constant of integration. However, since y(0) = 0, this implies that C = 0. Now we have
t*y(t) = t^2 + (K-1/3)*t^3/2
Integrating t*y(t) from 0 to 1 gives t^3/3 + (K-1/3)*t^4/8 evaluated at t = 1 minus its value at t = 0, so that gives
K = 1/3 + (K-1/3)*1/8
which has the unique solution K = 1/3. This in turn gives us our final answer:
y(x) = x.
No need for matlab or numerical approximations.

Torsten on 30 Mar 2015
i=0:
(y(1/2)-y(0))/(1/2)=1-1/3*0+0*integral_0^1(t*y(t))dt
-> 2*y(1/2)=2*y(0)+1-1/3*0+0*integral_0^1(t*y(t))dt
-> 2*y(1/2)=1
-> y(1/2)=1/2
Now do the same for i=1, and you are done.
Best wishes
Torsten.
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Roger Stafford on 4 Apr 2015
If you use the trapezoidal approximation to the integral, your exact solution will not quite satisfy your equation. Only if you use an exact integral using 'int' or calculus methods will the equation hold true.