Theoretical question on cyclic function

nhrock3
Messages
403
Reaction score
0
a(x) is continues on R with cycle T ,a(x+T)=a(x)
u(x) is non trivial soluion of y'=a(x)y
\lambda=\int_{0}^{T}a(x)dx

which of the following claims is correct:

A. if \lambda>0 then \lim_{x\rightarrow\infty}u(x)=\infty
B. if \lambda=0 then u(x) is a cyclic function

i don't have the theorectical basis to solve it
 
Physics news on Phys.org
The equation is separable. Write it as

\frac{y'}{y} = a(x)

and integrate both sides over one period.
 
how to integrate over a cycle
?
what to do after i integrate over a cycle
\int\frac{dy}{y}=\int a(x)dx

\ln y=\inta(x)dx
 
nhrock3 said:
how to integrate over a cycle
?
what to do after i integrate over a cycle
\int\frac{dy}{y}=\int a(x)dx

\ln y=\inta(x)dx

Try

\int_{u(0)}^{u(T)}\frac{dy}{y}=\int_0^T a(x)dx
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top