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Homework Help: Integral equation

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data

    solve the following integral equation

    [tex]y(x)=sin(x)+\int_0^\pi sin(x-t)y(t)dt[/tex]

    2. Relevant equations

    3. The attempt at a solution
    If the limits of integration were from 0 to x then i could solve this using Laplace transfroms because the definition of the convolution is:
    integral g(x-t)y(t) where the limits are from 0 to x.
    But here, we have the limits from 0 to pi.
    Does anyone have any ideas.
    Thank you
  2. jcsd
  3. Jan 6, 2010 #2


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    How about if you differentiate with respect to x twice. What happens if you add that to the original equation?
  4. Jan 6, 2010 #3
    after I differentiate once, I get:

    [tex]y'(x)=cos(x)(1+\int^\pi_0 (cos(t)-sin(t))y(t)dt)[/tex]

    differenitiating agin gives


    adding to the original gives

    [tex]y''+y=\int^\pi_0 (sin(x)-cos(x))sin(t)y(t)dt[/tex]

    is that right?
    What shall i do now?
  5. Jan 6, 2010 #4
    [tex]y(x)=sin(x)+\int_0^\pi sin(x-t)y(t)dt[/tex]
    let's try to solve this integral by integration by parts twice:
    [tex]\int_0^pi \sin(x-t)y(t)dt[/tex]
    let u=y(t) then du=y(t)dt
    dv=sin(x-t)dt then v=cos(x-t)
    apply [tex] u*v -\int vdu [/tex]
    we will obtain two factors [tex]A+\int cos(x-t)y(t)dt[/tex]
    another time integrate the second integral by parts we will obtain the same integral as above we will get 2*the first integral and solve for it!
  6. Jan 6, 2010 #5
    the first integral i wrote above i mean pi and not from 0--->p, writing mistake!
  7. Jan 6, 2010 #6
    I think you made a mistake,

    if u=y(t) then u'=y'(t) because we are integrating with respect to t.
    so in the intgrand we will have y'(t) and if we integrate by parts again, we will have y''(t) in the integrand.
  8. Jan 6, 2010 #7
    yes you'r right i have made a mistake.
    take u=y(t) du=y'(t)dt
    dv=sin(x-t)dt v=-cos(x-t)
    then after calculating it, calculate the second integral by the same method but take
    u=cos(x-t) du=sin(x-t)dt
    dv=y'(t)dt v=y(t)
  9. Jan 6, 2010 #8
    I dont think this is the right approach. after the integration by parts, we get that the integral =0 leaving us with
    y(x) = sin(x)

    is this right?
  10. Jan 6, 2010 #9


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    I get that y''(x)=(-sin(x))+integral(-sin(x-t)*y(t)). So y(x)+y''(x)=0. I don't see how you are factoring a cos(x) out of that first derivative.
  11. Jan 6, 2010 #10


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    How did you pull the cos(x) out of the integral? You just want to differentiate sin(x-t) wrt x and leave it all inside the integral, i.e.

    [tex]\int sin(x-t)y(t)dt \Rightarrow \int cos(x-t)y(t)dt[/tex]
  12. Jan 6, 2010 #11
    thank you both for pointing out my mistake.
    after differentiating twice and adding y and y'' I get
    so the solution is

    to find A and B I need initial conditions.
    How can I get the initial conditions?
  13. Jan 6, 2010 #12


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    Plug your y(x) into the integral equation. Now you can explicitly do the integral. If you move all of the terms to one side you will get something like c*sin(x)+d*cos(x)=0 (where c and d are linear functions of A and B). Now what?
  14. Jan 6, 2010 #13
    i did this just now and got that:



    is this right?
  15. Jan 6, 2010 #14


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    Right. You set c=d=0. Those look like the same values I got for A and B.
  16. Jan 6, 2010 #15
    wohoo, thanks.
    Im doing another question and it is the same as this one but instead of sin(x-t)y(t) in the integrand, i have (x-t)y(t) in the integrand.
    Do I also differentiate twice?
  17. Jan 6, 2010 #16


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    It's certainly worth a try. What do you think will happen?
  18. Jan 6, 2010 #17
    Why integration by parts is wrong in such a case?
  19. Jan 6, 2010 #18
    we have

    [tex]y(x)=1+\int_0^1 (x-t)y(t)dt[/tex]

    [tex]y'(x) = \int_0^1 y(t)dt[/tex]


    hmmm...do i add them? :confused:
  20. Jan 6, 2010 #19


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    Well, you don't need to. y''=0 is already an ODE you can easily solve.
  21. Jan 6, 2010 #20


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    For one thing, you did it wrong. If u=y then du=y'(x)dx. Now there's a y'(x) in your integral. The next time you integrate by parts you get y''(x). You aren't going to get back where you started. You are just making things more complicated.
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