Any idea how to solve equation?

[math] \int_0^1 C(yW(x))W^3(x)dx=F(y)[/math]

[math]W= 0.5(\cosh(kx)-cos(kx)-\frac{\cosh(k)+\cos(k)}{\sinh(k)+sin(k)}(\sinh(kx)-sin(kx)))[/math]

 

Need to find C

F(y) is known function.

k- is known constant

 

 

I don't understand.

 

How can F(y) be a function?

 

The integral (if it exists, which I haven't checked) is a definite integral and therefore a pure number and therefore a constant, not a function at all.

 

and what is y anyway?

Edited July 18, 20169 yr by studiot

Step one: simplify.

 

I'm not entirely sure what that would look like, I'm thinking something like this:

C (yW*4(x*3))d

 

W*2 would be W squared, but I don't know how to make the number smaller on my tablet. Also, this is probably wrong considering I don't do equations like this ever lol

[math]C( \xi )[/math] - is function of single variable.

 

This equation look like Fredholm equation of the first type.

But it can't be represent in form

[math]f(y)=\int_a^b K(x,y)\varphi(x)dx[/math]

where [math]K(x,t),f(x) [/math] is known functions and need to find [math] \varphi[/math]

 

In my case the [math]C(yW(x))=K(x,y)[/math] is not known and need to find.

 

I can replace variable

[math]\xi=yW(x) => F(y)=\frac{1}{y^4} \int_0^y C(\xi) \frac{\xi^3}{W'(x)} d\xi [/math], but

it means that I need to find relation of [math] x[/math] as function of variable ([math]\xi,y[/math] ) [math]x=\Phi(\xi,y)[/math]

it is impossible (look to definition of [math] W(x)[/math])

Soo we get Volterra equation of first kind

[math]F(y)y^4=\int_0^y C(\xi) K(\xi,y) d\xi[/math]

where [math]K(\xi,y)=\frac{\xi^3}{W'(\Phi(y,\xi))}, x=\Phi(\xi,y) [/math]- solution of equation [math]yW(x)=\xi [/math]

 

Edited July 19, 20169 yr by Airat

I don't understand.

 

How can F(y) be a function?

 

The integral (if it exists, which I haven't checked) is a definite integral and therefore a pure number and therefore a constant, not a function at all.

 

and what is y anyway?

 

The integration is with respect to x. F is a function of the variable y.

 

The integration is with respect to x. F is a function of the variable y.

 

Many thanks for your explanation.

 

I had understood that the integration was with respect to x.

I think I missed the first y.