Differential Equations and Linear Algebra Q&A

Can these equations be considered as differential equations?
Consider a differential equation with a term containing $y(x_0)$, for example
$$y^{″}-<!--\; -\; -->2y^{\prime }+y=y(x_0)$$
$x_0\in <!--\; \in \; -->\mathbb{R}$ is a constant. My question is, does such equations fall under the category of differential equations? I have never studied any equation with such a term. If its a differential equation, then $y(x_0)$ can be considered a constant coefficient?

Solving differential equations "c" value
I have two questions regarding solving differential equations given initial conditions:
1) When do you substitute the initial conditions into the equation to calculate the value of the constant "c". Do you substitute it once you integrate both sides of the differential equations and you get a constant "c"? Or do you substitute the initial conditions after integrating both sides AS WELL AS rearranging the equations to get y in terms of x and c. Using the second method, sometimes you get two values for "c" with only one value being correct.
2) When you solve certain differential equations, you get one side written with "$\pm <!--\; \pm \; -->$" in the front. However, only one equation fits the initial conditions even after you solve for the constant "c". The one that fits is either the one with the "+" or the one with the "−" in the front. How do you justify which one is correct without giving geometric representations of both and then saying "according to graph, this one insert equation is correct". Can you somehow solve without getting the "$\pm <!--\; \pm \; -->$" in the front?

Power series solution for differential equations
Is the solution to differential equations using power series applicable only to homogeneous differential equations? I mean equations of the form:
$$a_2{\mathrm{\varphi <!--\; \varphi \; -->}}^{″}(x)+a_1{\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }(x)+\mathrm{\varphi <!--\; \varphi \; -->}(x)=0$$

Decouple differential equations
I have a system of two Second Order differential equations
$$r^2\stackrel{¨<!--\; ¨\; -->}{r}-<!--\; -\; -->r^3({\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\phi <!--\; \phi \; -->}}}^2+{\mathrm{\omega <!--\; \omega \; -->}}^2)=-<!--\; -\; -->GM$$
$$r\stackrel{¨<!--\; ¨\; -->}{\mathrm{\phi <!--\; \phi \; -->}}+2\stackrel{\dot{}<!--\; \dot{}\; -->}{r}(\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\phi <!--\; \phi \; -->}}+\mathrm{\omega <!--\; \omega \; -->})=0$$
which I am supposed to decouple using the conservation size $(\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\phi <!--\; \phi \; -->}}+\mathrm{\omega <!--\; \omega \; -->})r^2$ I have shown, that it is indeed a conservation size, as its derivation is r-times the second equation and therefore zero. However I don't know how this is supposed to help me decoupling the two equations.

associated homogeneous linear differential equations
Can someone please explain how associated homogeneous linear differential equations work with an example?

Euler differential equations
I have following three equations
$$u^{″}-<!--\; -\; -->2u=-<!--\; -\; -->2v$$
$$u(0)=0$$
$$u^{\prime }(1)=0$$
and from these 3 equations I am trying to find u(v).
It looks to me "Cauchy-Euler Differential Equations - Nonhomogeneous case" but I am not sure about that because it is not an exactly Cauchy form. Could you help me to figure out u(v)?

differential equations of second order
How may I solve this differential equations:
$$y^{″}+4y=12x^2-<!--\; -\; -->16x\cos <!--\; \; -->(2x)?$$

Help with coupled differential equations
I have these two coupled differential equations
$$a\stackrel{¨<!--\; ¨\; -->}{x}+b\stackrel{⃛<!--\; ⃛\; -->}{x}+c\tan <!--\; \; -->(x+y)\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=0,$$
$$a\stackrel{¨<!--\; ¨\; -->}{y}-<!--\; -\; -->b\stackrel{⃛<!--\; ⃛\; -->}{y}-<!--\; -\; -->c\tan <!--\; \; -->(x+y)\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=0$$
with $a,b,c=const.$
Now I basically have no experience at all with more advanced differential equations. Are these even solvable? I guess answering this type of question is already a science to itself, but maybe you can share some intuition under which circumstances one can hope to find a solution? What are possible methods or tricks?

Find a particular solution to
$$y^{″}-<!--\; -\; -->4y^{\prime }+4y=\frac{14e^{3t}}{t^2+1}$$

what's the solution of the differential equation $y^{\prime }=\sqrt{(2x-<!--\; -\; -->y+1)}+2$

Show that ${\displaystyle y=\frac{2}{3}{e}^x+e^{-2x}}$ is a solution of the differential equation ${\displaystyle y^{\prime }+2y=2e^x}$

Solve the given differential equation by separation of variables.
${\displaystyle \csc \left(y\right)dx+{\sec }^2\left(x\right)dy=0}$

What would be the derivative of square roots? For example if I have ${\displaystyle 2\sqrt{x}}$ or ${\displaystyle \sqrt{x}}$

I have it down to
${\displaystyle \ln \left(y\right)+\left(\frac{x}{y}\right)y^{\prime }=y^{\prime }\ln \left(x\right)+\frac{y}{x}}$
The problem is factoring out y', which leads to either
${\displaystyle y^{\prime }=\frac{\ln \left(y\right)-\left(\frac{y}{x}\right)}{\ln \left(x\right)-\left(\frac{x}{y}\right)}}$
or to
${\displaystyle y^{\prime }=\frac{\left(\frac{y}{x}\right)-\ln \left(y\right)}{\left(\frac{x}{y}\right)-\ln \left(x\right)}}$
Am I missing something?

How to distinguish linear differential equations from nonlinear ones? I know, that e.g.:
${\displaystyle y{}^{″}-2y=\ln \left(x\right)}$
is linear, but
${\displaystyle 3+y{y}^{\prime }=x-y}$
is nonlinear. Why?

Find the general solution of the given differential equation.y'''-3y”+3y'-y=0

Find a differential equation whose general solution is ${\displaystyle y=c_1{e}^{-2t}+c_{2}t{e}^{-2t}}$

If ${\displaystyle x^2+y^2+z^2=9}$, dx / dt = 5, and dy /d1 = 4, find dz/ dt when (x, y, z) = (2, 2, 1).

Find two positive numbers whose product is 100 and whose sum is a minimum.

Find the general solution of the given differential equation.
${\displaystyle y{}^{″}+y=\tan }$
${\displaystyle 0<t<\frac{\pi }{2}}$