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methavix · 2007-01-19 20:13:51

hi all, i have the following integral equation to solve:

y x
/ /
| [1/(c1-c2*y)]dy = | [1/(1-x^2/c3^2)]dx
/ /
0 0

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?

Thanks
Luca

Chris · 2007-01-20 04:42:18

A slightly easier to read version:

$\int_0^y(c_1-c_2y)^{-1}dy = \int_0^x(\frac{c_3^2}{1-x^2})dx$

methavix · 2007-01-20 16:10:13

thanks but in the second integral the function should be:

c3^2/(c3^2-x^2)

Luca

BenTheMan · 2007-01-20 22:08:38

This is an interesting question. If you can get to a library, check for a book by Arfken and Weber called "Mathematical Methods in Physics" or something like taht---they cover integral equations in some detail.

Other than that I'm affraid that I don't have much experience with integral equations. I seem to remember that you can write them as differential equations, which are easier to solve.

Let me know how it works out.

$\int_0^y dy \frac{1}{c_1 - c_2y} = \int_0^x dx \frac{c_3^2}{c_3^2 - x^2}$

Bishop · 2007-01-21 03:22:22

There might be some intracies that I'm missing out on here, but I'm just goint to blindly go ahead anways... Assuming that the *real* equation has little prime ticks inside the integration, i.e. dummy variables, i.e.

$\int_0^y dy' \frac{1}{c_1 - c_2y'} = \int_0^x dx' \frac{c_3^2}{c_3^2 - x'^2}$

then both sides of the equation are pretty easily integrated. This gives us just a regular ol' equation (no derivates) which is

$\frac{1}{c_2}ln[\frac{c_1}{(c_1-c_{2y})}] = \frac{c_3}{2}ln[\frac{(c3+x)}{(c3-x)}]$ .

Solving this equation give us the expression for y(x) as

$y(x) = \frac{c_1}{c_2}(1-(\frac{c_3+x}{c_3-x})^{-0.5*c_2c_3})$ .

Yeah... It does seem like an integral equation should have some of the complexities of a differential equation, but this equation just struck me as one that is much more simple than how it was presented. That is, it looked exactly solveable, with all the complex integrals put in just to confuse the situation. Is it this simple? Did I make sense?