Calculus 2 Equations: Expert Solutions and More

How to prove this equality with closed form?
An exercise I'm trying to solve gives us
${\displaystyle F\left(y\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(xy\right)}{x(1+x^2)} dx }$
and first asks us to prove that F(y) satisfies the ODE ${\displaystyle F{ }^{″}\left(y\right)-F\left(y\right)+\frac{\pi }{2}=0}$, then find the closed form of the ODE and lastly prove that:
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\cos \left(yx\right)}{1+x^2} dx =\frac{\pi e^{-y}}{2}}$
using the closed form of the ODE.
I've gone through the first two and for the closed form I got ${\displaystyle F\left(y\right)=c_1{e}^{-y}+c_2{e}^y+\frac{\pi }{2}}$.
But I'm having trouble with solving the last part of the exercise. With complex analysis, I can solve it, but I don't know what to do with the closed form. I'm sharing my original thoughts, but I get stuck:
First, I noticed that the last integral is the derivative of F(y):
$F^{\text{'}}\left(y\right)={\int }_0^{\mathrm{\infty }}\frac{d}{dy}\left(\frac{\sin \left(xy\right)}{x(1+x^2)}\right)dx={\int }_0^{\mathrm{\infty }}\frac{\cos \left(yx\right)}{1+x^2}dx$
Also: ${\displaystyle F^{\prime }\left(y\right)=-c_1{e}^{-y}+c_2{e}^y}$. Then:
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\cos \left(yx\right)}{1+x^2} dx =-c_1{e}^{-y}+c_2{e}^y}$
Then I'd need to show ${\displaystyle -c_1{e}^{-y}+c_2{e}^y=\frac{\pi e^{-y}}{2}}$ and I get stuck here.
Am I doing something wrong? Is this even solvable this way? Should I approach it differently?

Find u such that
${\displaystyle -u^{{ }^{″}}\left(x\right)+b\left(x\right)u^{{ }^{\prime }}\left(x\right)+c\left(x\right)u\left(x\right)=f\left(x\right)\text{ in }(0,1),}$
and conditions ${\displaystyle u\left(0\right)=u\left(1\right)=0}$, where
${\displaystyle b\left(x\right)=x^2,c\left(x\right)=1+x,f\left(x\right)=-2+13x^2+3x^3-x^4-5x^5}$

Find a matrix such that the systems are equivalent
Take the system
$$\{\begin{array}{l}w=v^{\prime }\\ v^{″}-\mu v^{\prime }+v=0\end{array}$$
for ${\displaystyle \mu \in \mathbb{R}}$
Find a constant matrix ${\displaystyle A^{\mu }}$ such that the above system is equivalent to
$\left[\begin{array}{c}v\text{'}\\ w\text{'}\end{array}\right]=A^{\mu }\left[\begin{array}{c}v\text{'}\\ w\text{'}\end{array}\right]$
My first thought was to find a solution to ${\displaystyle v{ }^{″}-\mu v^{\prime }+v=0}$. Taking the characteristic polynomial as ${\displaystyle r^2-\mu r+1=0}$, I get ${\displaystyle r=\frac{1}{2}(\mu \pm \sqrt{{\mu }^2-4})}$. Not really sure what to do at this point, however. Another way to solve this might be first to replace v' in the second equation with w to get ${\displaystyle v{ }^{″}-\mu w+v=0}$ but I'm not sure where to go from there either.

Emergence of Cauchy Principal value
I have a problem solving differential equation, where I think there should be a Cauchy Principal value involved, but I do not see how it it should emerge. Let's say we have a differential equation:
$\mathrm{i}{\dot{C}}_1\left(t\right)=\underset{0}{\overset{\infty }{\int }}d\omega \nu \left(\omega \right)e^{\mathrm{i}({\omega }_{10}-\omega )t}$,
where all the variables are real and positive. If we now solve the differential equation, it needs only to carry out the time integral
$\mathrm{i}{C}_1\left(t\right)=\int dt\underset{0}{\overset{\infty }{\int }}d\omega \nu \left(\omega \right)e^{\mathrm{i}({\omega }_{10}-\omega )t}=\underset{0}{\overset{\infty }{\int }}d\omega \frac{\nu \left(\omega \right)}{\mathrm{i}({\omega }_{10}-\omega )}{e}^{\mathrm{i}({\omega }_{10}-\omega )t}$.
So the problem is that if the last integral is not taken as principal value, it will give ${\displaystyle \mathrm{\infty }}$, so I would rather need it to be
$\mathcal{P}\underset{0}{\overset{\infty }{\int }}d\omega \frac{\nu \left(\omega \right)}{\mathrm{i}({\omega }_{10}-\omega )}{e}^{\mathrm{i}({\omega }_{10}-\omega )t}$
where ${\displaystyle \mathcal{P}}$ is the principal value. Would anyone have any idea where the principal value would emerge from, if at all? I was thinking maybe it has something to do when swapping the order of integrals in order to carry out the time integral first, but I am not sure.

Solving system of differential equations : Wolfram Alpha vs theorem
I am burning my brain finding the most correct way to solve a system of differential equations. Here is an example :
$$\{\begin{array}{l}{x}^{\prime }=5x-2y\\ y^{\prime }=-x+6y\end{array}$$
Let's $Y\left(t\right)=\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)$. I use a matrix $A=\left(\begin{array}{cc}5& -2\\ -1& 6\end{array}\right)$, calculate its eigen values ${\displaystyle {\lambda }_1=7,{\lambda }_2=4}$ and eigen vectors $v_1=\left(\begin{array}{c}-1\\ 1\end{array}\right),v_2=\left(\begin{array}{c}2\\ 1\end{array}\right)$.
Now 2 options:
- I can use a theorem and find the solution is ${\displaystyle Y\left(t\right)=c_1{e}^{{\lambda }_{1}t}{u}_1+c_2{e}^{{\lambda }_{2}t}{u}_2}$
- I can continue using linear algebra, calculate P, D and ${\displaystyle P^{-1}}$ such that ${\displaystyle A=PD{P}^{-1}}$ where D is a diagonal matrix. ${\displaystyle Y^{\prime }=AY}$ so $Y=C{e}^{tA}=CP\left(\begin{array}{cc}{e}^{7t}& 0\\ 0& e^{4t}\end{array}\right)P^{-1}$
I don't find same results but both are OK (the second is used by Wolfram). Did I misunderstand something? Which one is the best option?

Solving a systemof eqns by given initial conditions
${\displaystyle x^{\prime }\left(t\right)=2x+8y}$
${\displaystyle y^{\prime }\left(t\right)=-x-2y}$
Which has the I.C. ${\displaystyle X\left(0\right)=(6,-2)}$. So I take that it means this:
${\displaystyle 6=2x+8y}$
${\displaystyle -2=-x-2y}$
and then it is solved as any other system? We get eigenvector ${\displaystyle (1,\frac{1}{2})}$, but since the eigenvalues of the system are two, ${\displaystyle \pm 2i}$, we have to find the second generalized eigenvector.
$\left(\begin{array}{cc}2-2i& 8\\ -1& -2-2i\end{array}\right)\left(\begin{array}{c}1\\ \frac{1}{2}\end{array}\right)=\left(\begin{array}{c}6-2i\\ -2-i\end{array}\right)$
So the second vector would be ${\displaystyle (6-2i,-2-i)}$. Using the general solutions for the system, we get
$y\left(t\right)=e^{2it}(1,\frac{1}{2})+e^{-2it}(6-2i,-2-i)$
Would this be correct, or did I misinterpret the initial conditions?