evinda said:
Hello! (Wave)
We have this: $y''+y=\frac{1}{\cos x} , y'(0)=0, y(\pi)=0$.
Using the Green function I got that $y(x)= x \sin x+ \cos x( -\ln |\cos \pi|+ \ln |\cos x|)=x \sin x+ \cos x (\ln |\cos x|)$.
But according to Wolfram:
y'''''''+'y'='1'/'cosx , y''''('0')''='0, y'('pi')''='0 - Wolfram|Alpha
How do we get the imaginary part?
Hey evinda! (Smile)
Wolfram always treats functions as complex functions.
That has in particular impact on the natural logarithm.
Consider that if we're talking about
real functions that we actually have:
$$
\int \frac{dx}{x} = \begin{cases}\ln(x) + C_1&\text{if }x>0 \\ \ln(-x) + C_2&\text{if }x<0\end{cases}
$$
That's because the domain is split into 2 disconnected intervals for which we have independent integration constants.
This is usually written as:
$$\int \frac{dx}{x} = \ln|x| + C$$
with the understanding that the integration constant $C$ can be different on each part of the domain. (Nerd)
However, with
complex functions there is no such distinction - all of the domain is connected with only an undefined point at $z=0$. That is:
$$
\int \frac{dz}{z} = \ln z + C
$$
Furthermore, $\ln z$ is a so called
multivalued function that behaves a bit different than a normal function.
That is, we can't just take the natural logarithm of a complex number. (Sweating)So instead of:
$$
-\ln|\cos\pi| + \ln|\cos x|
$$
we really have something like:
$$
-\ln(\cos\pi) + \ln(\cos x) = \ln\frac{\cos x}{\cos\pi} = \ln(-\cos x)
$$
Now let's define $z = \ln(-\cos x)$, so that we have $e^z = -\cos x$.
Since we're talking about complex numbers, we can write this as:
$$e^{z+2\pi ik} = -\cos x
\quad\Rightarrow\quad e^{z+i\pi+2\pi ik} = \cos x
\quad\Rightarrow\quad z+i\pi+2\pi ik = \ln(\cos x)
\quad\Rightarrow\quad z = \ln(\cos x)-i\pi-2\pi ik
$$
(Whew)