Volterra Eqn of 2nd Kind -> DEQ

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The discussion focuses on converting the Volterra Equation of the 2nd Kind, specifically y(x) = 1 - x + ∫[dt(x-t)y(t)] from 0 to x, into a differential equation (DEQ) with appropriate boundary conditions. The user correctly identifies the form of the DEQ as a 2nd order homogeneous equation, d²y/dx² + y(x) = 0, and establishes boundary conditions y(0) = 1 and dy/dx|0 = -1. The conversation highlights the importance of applying differentiation within the integral and correcting sign errors in the integral to achieve the correct formulation of the DEQ.

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Volterra Eqn of 2nd Kind --> DEQ

Homework Statement


I need to convert y(x) = 1 - x + int[dt(x-t)y(t)] from 0 to x to a differential equation with the appropriate boundary conditions.





The Attempt at a Solution



OK I just had a problem converting a DEQ into an integral equation so I know the form it will take, I know it will be a 2nd order homogeneous equation... something like d^2y/dx^2 + y(x) = 0 with y(a) = b , dy/dx|c = d for some constants. I know it won't be a periodic boundary condition because that turns into a Fredholm Equation of the 2nd Kind.

So I applied d/dx to both sides getting

dy/dx = -1 + int[dty(t)] from 0 to x

I think this is valid since the integrand is with respect to t, I just applied the differentiation within the integral.

I'm not sure what to do from here though.
 
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Ok so I missed the obvious step of solving for the boundary conditions by simply plugging in x= 0 into my y(x) and dy/dx equations

so y(0) = 1 and dy(0)/dx = -1

This is what I got from my sketchy inspection method prior to this. I know the answer now, but I still don't know how to show it will become a 2nd order homogeneous equation.
 
First off there are two terms in the derivative of the integral since there are two sources of x dependence in the integral (one in the function and one in the limit). The one you have ignored does indeed vanish - is that why you ignored it? Once you get to this:

dy/dx = -1 + int[dty(t)] from 0 to x

you have a sign error in the integral. After fixing it you will have a correct equation, but it's still an integral equation. Suggests that you might want to differentiate again, yes?
 

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