# What is the solution for the attached equation?

Good afternoon,
i was just wondering if this equation is possibly solvable where I(z) is a function of z. The equation is:
I(z)=cosh(1/2 ∫I(z)dz)
I know it looks stupid but is it possible? How would you approach this problem?
Thank you.

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Svein
Try to derive both side with respect to z. This will give you an ordinary differential equation. Then try to solve it in another way.

Thanks for the response. However I get hyperbolic sinh which just makes it much worse. :(

Samy_A
Maybe first set $$arcosh(I(z))=\frac {1}{2} \int_z^L I(x)dx$$ and then take the derivate?
If you introduce $$J(z) = \sinh\left(\frac12 \int I(z)\,dz\right)$$ then you can get the two-dimensional system $$I' = \frac12 IJ \\ J' = \frac12 I^2$$ which can be solved numerically subject to given initial conditions (which, like cosh and sinh, must satisfy $I(0)^2 - J(0)^2 = 1$). Alternatively, you can construct power series iteratively by starting with $I_0(z) = I(0)$, $J_0(z) = J(0)$ and using the recurrence relation $$I_{n+1}(z) = I(0) + \frac12 \int_0^z I_n(t)J_n(t)\,dt, \\ J_{n+1}(z) = J(0) + \frac12 \int_0^z I_n^2(t)\,dt.$$