Calculus 2 Equations: Expert Solutions and More

How to solve coupled second order differential equations
${\displaystyle 2y{}^{″}-3y^{\prime }+2z^{\prime }+3y+z=e^{2x}}$
${\displaystyle y{}^{″}-3y^{\prime }+z^{\prime }+2y-z=0}$

How to solve
${\displaystyle \ddot{\overrightarrow{u}}=\overrightarrow{u}\times \hat{k}}$

Proving single solution of initial value problem is increasing
Given the initial value problem
${\displaystyle y^{\prime }\left(x\right)=y\left(x\right)-\sin \left\{y\left(x\right)\right\},y\left(0\right)=1}$

Proving identity
${\displaystyle \sum [{(j+t)}^{-1}-j^{-1}]=\sum _{k\ge 1}\zeta (k+1){(-t)}^k}$

Differential equations: Shouldn't the method of separation of variables miss the solutions when ${\displaystyle x=0}$?
Consider the ODE:
${\displaystyle x\frac{dy}{dx}=2y}$ (1)
We apply the separation of variables technique to obtain:
${\displaystyle \frac{1}{y}dy=\frac{2}{x}dx}$ (2)
which provides the general solution:
${\displaystyle y\left(x\right)=C{x}^2}$
The book mentions that "the general solution (3) satisfies (1) for any value of the constant C and for all values of the variable x".
However, I believe the statement is wrong when ${\displaystyle x=0}$. By using the separation of variables, we divided by x and y to obtain (2). Thus, we make the implicit assumption that ${\displaystyle x\ne 0}$, and the separation of variables will generate solutions where ${\displaystyle x\ne 0}$.
How can y be defined for any x, when we implicitly assume that ${\displaystyle x\ne 0}$?

Determining convergence of a series
${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\sqrt{n}(\sqrt{n^4+1}-n^2)}$

Determine the general solution of the given differential equation:
${\displaystyle y{}^{″}-4y^{\prime }+13y=e^{2x}}$

Derivative of a differential equation
${\displaystyle y^{\prime }=x^2-y^2}$
then why does ${\displaystyle y{}^{″}=2x-2y{y}^{\prime }}$ shouldn't ${\displaystyle y{}^{″}=\frac{d}{dx}{y}^{\prime }=2x-2y}$
I am trying to understand ODE's, so maybe a simple question requiring a simple response. Where is the extra y' coming from on 2yy'

Convert unit measures for a problem from zill book
I have the solution, but I don't understand it, I don't understand the part which implies the conversion of unit measures. I know that I can ${\displaystyle g=32\cdot \frac{ft}{s^2}}$, but then what? So my questions are:
1. how to compute (convert unit measures) ${\displaystyle \frac{4pound}{32\cdot \frac{ft}{s^2}}}$ to become ${\displaystyle \frac{1}{8}}$?
pound is a unit measure for weight, ft is for distance (like meter), s is for time. So, how they (in image) simplify that raport - division?
2. Then, for the next step: I know that ${\displaystyle {\omega }^2=\frac{k}{m}}$, with ${\displaystyle m=\frac{1}{8}}$ slug and ${\displaystyle k=16\frac{lb}{ft}}$. But what about this division:
${\displaystyle \frac{k}{m}=\frac{16\frac{lb}{ft}}{\frac{1}{8}slug}=128\frac{lb\cdot slug}{ft}}$
And again, how can I make this division? lb and slug are units of measure for weight and ft is for distance. How can I reduce them?

Convergence of a series and respective sum:
${\displaystyle \sum _{n\ge 0}{\left(\frac{2}{3}\right)}^{2n-1}}$

Consider the Regular Sturm Liouville(RSLP) problem
${\displaystyle {\left(p\left(x\right)y^{\prime }\right)}^{\prime }+(q\left(x\right)+\lambda r\left(x\right))y=0}$

${\displaystyle (1-x^2)y{}^{″}-x{y}^{\prime }+m^{2}y=0}$, m is a constant. I have to show that ${\displaystyle m\in \mathbb{Z}\Rightarrow deg\left(y\right)=m}$ such that y is a solution of the ode.

Differential form of a differential equation
By multiplying the differential equation
${\displaystyle 6xy\frac{dy}{dx}+x^2+y^2=0}$
by dx we see that the equation has the alternative differential form
${\displaystyle (x^2+y^2)dx+6xydy=0}$.

Determining the sign of a solution to a Cauchy problem
Suppose that ${\displaystyle y\in C^1\left(\mathbb{R}\right)}$ is the solution of
$\{\begin{array}{l}{y}^{\prime }(x)=\sin (y(x)+x^2)x\in \mathbb{R},\\ y(0)=0.\end{array}$

Differential Equations, Undetermined Coefficients - Equal Terms in the Homogeneous and Particular Solutions
${\displaystyle y{}^{″}-4y^{\prime }+3y=\text{cosh}\left\{x\right\}=\frac{e^x+e^{-x}}{2}}$

Conflicting results in finding orthogonal curves
I'm told to find the set of orthogonal curves to the curve of equation ${\displaystyle y=c{x}^2}$. Using implicit differentiation, ${\displaystyle \frac{dy}{dx}=2cx}$. Then, the desired curves obey the differential equation ${\displaystyle \frac{dy}{dx}=-\frac{1}{2cx}}$. Integrating, we find the curves ${\displaystyle y=-\frac{1}{2c}\log \left\{\left|x\right|\right\}+K}$, for some real constant K. The thing is in some solution I found, they rewrite 2cx as ${\displaystyle \frac{2y}{x}}$, because ${\displaystyle c=\frac{y}{x^2}}$, and in doing so they get a different differential equation with different curves ${\displaystyle (y=\pm \sqrt{K-\frac{1}{2}{x}^2})}$. Which are clearly different curves. Plotting them on geogebra makes it seem like mine is wrong. What error did I make?