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Integral equation

  1. Jan 19, 2007 #1
    hi all, i have the following integral equation to solve:

    | [1/(c1-c2*y)]dy = | [1/(1-x^2/c3^2)]dx

    where c1, c2, c3 are constants.

    Can I solve it analytically? If not, how can i dot to find the function x(y) or y(x)? Otherwise, does it exist an approximated way to find the analytical solution? Can I solve it only numerically?

  2. jcsd
  3. Jan 19, 2007 #2


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    I think maybe you are being confused because the x and y are appearing both as limits of integration and as the variables of integration. The variables of integration are dummy symbols. If you just perform the integrations its then easy enough to solve for y(x).
  4. Jan 19, 2007 #3


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    In other words, treat it as
    [tex]\int_0^y \frac{1}{c_1- c_2t}dt= \int_0^x \frac{1}{1- \frac{s^2}{c_3^2}} ds[/tex]

    If what you have written is really the problem, then the left side of the equation is a function of y only and the right hand side is a function of x only. The only way they can be equal is if they are each equal to the same constant:
    [tex]\int_0^y \frac{1}{c_1- c_2t}dt= A[/itex]
    [tex]\int_0^y \frac{1}{1- \frac{s^2}{c_3^2}}ds= A[/tex]

    However, as Dick said, those aren't really "integral equations"- there is no variable occuring inside the integral other than the dummy "variable of integration". In particular, those two integrals are NOT constant and so can't be equal! Something doesn't make sense.

    Editted after looking at benorin's solution: My last paragraph is wrong because y is a function of x- what I said is only true if x and y are independent variables. Sorry.
    Last edited by a moderator: Jan 20, 2007
  5. Jan 20, 2007 #4


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    For the lefthand-side, substitute [tex]u=c_1-c_2Y\Rightarrow -c_2dY[/tex] so that [tex]0\leq Y\leq y\Rightarrow c_1\leq u\leq c_1-c_2y[/tex] and the integral becomes

    [tex]\int_{0}^{y}\frac{dY}{c_1-c_2Y}=-\frac{1}{c_2}\int_{c_1}^{c_1-c_2y}\frac{du}{u}=-\frac{1}{c_2}\left[\ln |u|\right]_{u=c_1}^{c_1-c_2y}=\frac{1}{c_2}\ln\left|\frac{c_1}{c_1-c_2y}\right|[/tex]​

    and for the righthand-side substitute [tex]u=\frac{X}{c_3} \Rightarrow \frac{1}{c_3}dX[/tex] so that [tex]0\leq X\leq x\Rightarrow 0\leq u\leq \frac{x}{c_3}[/tex] and the integral becomes


    using partial fractions we have

    [tex]=\frac{c_3}{2}\int_{0}^{\frac{x}{c_3}}\left(\frac{1}{u+1}-\frac{1}{u-1}\right) du=\frac{c_3}{2}\left[\ln |u+1|-\ln |u-1|\right]_{0}^{\frac{x}{c_3}} =\frac{c_3}{2}\left[\ln\left|\frac{u+1}{u-1}\right|\right]_{0}^{\frac{x}{c_3}}[/tex]
    [tex]=\frac{c_3}{2}\ln\left|\frac{\frac{x}{c_3}+1}{\frac{x}{c_3}-1}\right|-\ln 1 = \frac{c_3}{2}\ln\left|\frac{x+c_3}{x-c_3}\right|[/tex]​

    hence the final equation is

    [tex]\frac{1}{c_2}\ln\left|\frac{c_1}{c_1-c_2y}\right|= \frac{c_3}{2}\ln\left|\frac{x+c_3}{x-c_3}\right|[/tex]​

    I will solve for $y(x)$ (a little easier):

    [tex]\ln\left|\frac{c_1}{c_1-c_2y}\right|= \frac{c_2c_3}{2}\ln\left|\frac{x+c_3}{x-c_3}\right|= \ln\left|\frac{x+c_3}{x-c_3}\right|^{\frac{c_2c_3}{2}}[/tex]



    [tex]c_1-c_2y=\pm |c_1|\left|\frac{x+c_3}{x-c_3}\right| ^{-\frac{1}{2}c_2c_3} [/tex]

    [tex]\boxed{y=\frac{c_1\mp|c_1|\left|\frac{x+c_3}{x-c_3}\right|^{-\frac{1}{2}c_2c_3}}{c_2} }[/tex]​
    Last edited: Jan 20, 2007
  6. Jan 20, 2007 #5
    thank you very much
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