@HallsofIvy: Thanks for taking the time and effort to write your answer! I really appreciate it. However, you assumed erroneously... My integral equation is a Voltera of the second kind... The thing is that for a seperable kernel with one term (like the one I am looking for) the solution is...
Yes, I have tried that. However, then the integral equation is not solvable, while if the kernel can be written with only one term as K(x,a)=A(a) X(x) then there exists an analytical solution...
Hi,
I need some help factorizing the following:
\frac{g(x)-g(a)}{u(a)^2}=K(x,a)
into A(a) X(x)
ie I want to find A(a), X(x) such that their product is the first equation. The reason I want to do this is because K(x,a) is a kernel and it would help a lot if somehow I could write it as...
Actually, setting:
Q(t)=e^{\int_0^t dx G(x)}
reduces your 2nd order ODE to a 1st order nonlinear one:
G'(t)=-G(t)^2+f(t) G(t)+g(t)+h(t) \sqrt{G(t)^2 k(t)^2+1}
Depending on the form of the functions f(t),h(t), g(t) and k(t) you might be able to solve the latter and then again it...
Hmmm, when I said eqs. (1) & (2) I obviously meant form my first post... The point is to transform:
y''(x)+P(x) y'(x)+Q(x) y(x)=0...Equation (1)
into
f''(x)+P(x) f'(x)+Q(x)/x f(x)=0...Equation (2)
given the P(x) & Q(x), where:
P(x)=\frac{3}{x}+\frac{H'(x)}{H(x)}
and...
I bet it does! I was just wondering if it can be done for arbitrary functions... However, here you go:
P(x)=\frac{3}{x}+\frac{H'[x]}{H[x]}
Q(x)=\frac{c}{x^5 H(x)^2}
Again, H(x) is arbitrary... sort of... Nevertheless, it is easy to see that eq (2) with these definitions is solvable (for...
Hi, I have a problem and I am wondering if anyone can help... There is
this ODE, where P(x) and Q(x) are known functions:
y''(x)+P(x) y'(x)+Q(x) y(x)=0 (1)
This ODE cannot be solved analytically in general. However I can solve the following one (for the specific P(x) and Q(x) I have...