Airat Posted July 18, 2016 Share Posted July 18, 2016 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 Link to comment Share on other sites More sharing options...
studiot Posted July 18, 2016 Share Posted July 18, 2016 (edited) 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, 2016 by studiot Link to comment Share on other sites More sharing options...
Raider5678 Posted July 19, 2016 Share Posted July 19, 2016 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 Link to comment Share on other sites More sharing options...
Airat Posted July 19, 2016 Author Share Posted July 19, 2016 (edited) [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, 2016 by Airat Link to comment Share on other sites More sharing options...
Country Boy Posted August 20, 2016 Share Posted August 20, 2016 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. Link to comment Share on other sites More sharing options...
studiot Posted August 20, 2016 Share Posted August 20, 2016 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. Link to comment Share on other sites More sharing options...
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