• Support PF! Buy your school textbooks, materials and every day products Here!

Integral equation

  • Thread starter Methavix
  • Start date
38
1
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
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
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).
 
HallsofIvy
Science Advisor
Homework Helper
41,736
894
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]
and
[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:
benorin
Homework Helper
1,067
14
hi all, i have the following integral equation to solve:

[tex]\int_{0}^{y}\frac{dY}{c_1-c_2Y}=\int_{0}^{x}\frac{dX}{1-\frac{X^2}{c_3^2}}[/tex]​

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
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

[tex]\int_{0}^{x}\frac{dX}{1-\frac{X^2}{c_3^2}}=c_3\int_{0}^{\frac{x}{c_3}}\frac{du}{1-u^2}[/tex]​

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

[tex]\left|c_1-c_2y\right|=|c_1|\left|\frac{x+c_3}{x-c_3}\right|^{-\frac{1}{2}c_2c_3}[/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:
38
1
thank you very much
 

Related Threads for: Integral equation

  • Last Post
Replies
13
Views
994
  • Last Post
Replies
0
Views
828
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
2
Replies
32
Views
3K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
515
Top