Solving an integral equation by iteration

1. Jan 31, 2013

mode1111

Hello guys!

I was given a Volterra integral equation y(x)=1/2*x^2+integral(0--->x) [t(t-x)y(t)]dt to solve using iteration. I have no idea how and where to start...

The full problem goes as follows:
Show that the solution y(x) of y''+xy=1, y(0)=y'(0)=1 also satisfies the integral equation (above). Use iteration to solve the integral equation, and compare with the series solution of the differential equation.

The series solution (done with taylor) equals: 1+x+1/2*x^2-1/2*x^3-1/6*x^4-.....
when you turn the differential equation to an integral equation it looks the same except it has additional 1+x that come from the initial conditions. So basically, to compare them we need different initial conditions i.e y(0)=y'(0)=0 and not 1. However I've asked my professor, and he said there is no such problem and that the algebra was wrong. I can't find the error...

Thanks to anyone that tries!

2. Jan 31, 2013

HallsofIvy

I don't understand why you would need different initial conditions. You are comparing different solutions to the same initial value problem aren't you? Changing the initial conditions would change the problem.

And why do you want to "turn the differential equation to an integral equation"? You are given the integral equation to begin with.

3. Jan 31, 2013

mode1111

I totally agree with you, however if you try to iterate the integral equation and to solve the differential one using series solution, you'll find that you lack the 1+x part. This fact made me question the initial conditions so i tried to make the two equations look the same. It turned out that they are indeed the same, but the conditions are different, i.e the integral one doesn't have the 1+x part that comes from y(0)=y'(0)=1. Again, I did something wrong here and I don't know what. If you could please explain or write down your solution, I'll be more than grateful to you...
Thank you.

4. Feb 1, 2013

HallsofIvy

Well, this can't be right. From the integral equation, $y(0)= (1/2)(0)+ \int_0^0 [t^2y(t)dt= 0$, not 1. Further, $y'(x)= x- \int_0^x ty(t)dt$ so that $y(0)= 0- \int_0^0 ty(t)dt= 0$. The differential equation is correct but the initial values are wrong. They should be y(0)= y'(0)= 0.

5. Feb 1, 2013

mode1111

Yeah, that is exactly my point...I guess the my professor was wrong after all.
Thanks, HallsofIvy!