# Solving an Integro-differential equation numerically

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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

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### Answers (3)

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.

##### 0 Comments

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.

##### 2 Comments

Roger Stafford
on 4 Apr 2015

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