Differential Equations and Linear Algebra Q&A

Deriving Stochastic differential equations. I am having a difficulty in deriving stochastic differential equations from geometric Brownian motion dynamics.
Assume S follows the geometric Brownian motion dynamics, $dS=\mathrm{\mu <!--\; \mu \; -->}Sdt+\mathrm{\sigma <!--\; \sigma \; -->}SdZ$, with $\mathrm{\mu <!--\; \mu \; -->}$ and $\mathrm{\sigma <!--\; \sigma \; -->}$ constants. Derive the stochastic differential equation satisfied by $y=2S,y=S^2,y=e^S$

Differential equations notations confusion
Given these differential equations:
$\frac{d^{2}x}{d{t}^2}=2\mathrm{\Omega <!--\; \Omega \; -->}\frac{dy}{dt}\sin <!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})-<!--\; -\; -->\frac{g}{L}x$
$\frac{d^{2}y}{d{t}^2}=-<!--\; -\; -->2\mathrm{\Omega <!--\; \Omega \; -->}\frac{dx}{dt}\sin <!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})-<!--\; -\; -->\frac{g}{L}y$
Now making the following substitutions:
$\frac{dx}{dt}=u$ and $\frac{dy}{dt}=v$ we have the following differential equations
$\frac{du}{dt}=2\mathrm{\Omega <!--\; \Omega \; -->}v\sin <!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})-<!--\; -\; -->\frac{g}{L}x$
$\frac{dv}{dt}=-<!--\; -\; -->2\mathrm{\Omega <!--\; \Omega \; -->}u\sin <!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})-<!--\; -\; -->\frac{g}{L}y$
Now here is where I get confused, note that we can write $\frac{du}{dt}=f(t,u)$ and likewise for the other equation. But if that is so, then what would be my unknown variable for the first equation? Or is it wrong to write $\frac{du}{dt}=f(t,u)$?

Differential equations in function
Equations (1): $x{y}^{\prime }+(1-<!--\; -\; -->x)y=1$
determine and solve the differential equation (2) whose general solution is the function z .
determine the general solution of (1)

Getting into differential equations
I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:
$$u^{\prime }+u=0,y(x)=C{e}^{-<!--\; -\; -->x}$$
How do I tackle this?

Is this system of differential equations linear?
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=x+y+5$
$\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=x-<!--\; -\; -->y$
Is this system linear or nonlinear differential equations?

Set of coupled partial differential equations
I've read that the Einstein equation is a set of 10 coupled partial differential equations. I know what a partial differential equation is, but I don't know what a set of coupled partial differential equations is. Please shed some light into this question.

System of differential equations
$x^{\prime }=4x-<!--\; -\; -->2y$
$y^{\prime }=3x-<!--\; -\; -->y-<!--\; -\; -->2e^{3t}$
Initial conditions are $x(0)=y(0)=0$

Multiplied differential equations
How to get x(t),y(t) solutions for "product differential equations" (dotted on t):
$$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=xy,{\stackrel{\dot{}<!--\; \dot{}\; -->}{y}}^2-<!--\; -\; -->{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2=1;$$
we have by solving quadratics
$$(2{\stackrel{\dot{}<!--\; \dot{}\; -->}{y}}^2,2{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2)=+1\pm <!--\; \pm \; -->\sqrt{1+4x^2{y}^2},-<!--\; -\; -->1\pm <!--\; \pm \; -->\sqrt{1+4x^2{y}^2}$$

Differential Equations Integrating y by x
This may be a bit of a silly questions, but when solving a differential equation by finding an integrating factor, is it possible to integrate a function of y and x by x? I understand that in multi variable calculus the y would be treated as a constant, but I am not sure why the same does not apply in differential equations. For example...
For the differential equation $y^{\prime }+y=x{y}^3$
The way to solve it would be to multiply the whole equation by $-<!--\; -\; -->2y^{-<!--\; -\; -->3}$ then solve for the integrating factor which would be $e^{-<!--\; -\; -->2x}$.
But I'm wondering why we even need to get rid of the y terms on the right side of the equations. Can we solve the equations as such...
$y^{\prime }+y=x{y}^3$
$y^{\prime }\ast <!--\; \ast \; -->e^x+e^{x}y=x{y}^3{e}^x$
$\int <!--\; \int \; -->d(y{e}^x)=\int <!--\; \int \; -->x{y}^3{e}^{x}dx$
I understand that this method is incorrect, but I am having trouble understanding why separation of variables are absolutely necessary in differential equations.

Simple Differential Equations
I want to solve the two following differential equations:
(1) $f^{″}(t)=3f^{\prime }(t)-<!--\; -\; -->f(t)$
(2) $f^{″}(t)=2f^{\prime }(t)-<!--\; -\; -->f(t)$
I chose the approach $f(t)=e^{\mathrm{\lambda <!--\; \lambda \; -->}t}$ and hence arrive for the first case at
${\mathrm{\lambda <!--\; \lambda \; -->}}^2{e}^{\mathrm{\lambda <!--\; \lambda \; -->}t}=e^{\mathrm{\lambda <!--\; \lambda \; -->}t}(3\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->1)\to <!--\; \to \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^2=3\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->1$
and for the second differential equation I arrive at ${\mathrm{\lambda <!--\; \lambda \; -->}}^2=2\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->1$
Is this correct? How do I find $\mathrm{\lambda <!--\; \lambda \; -->}$ now per hand easily? Also, in the exercise it says "find the solution and determine what $\mathrm{\lambda <!--\; \lambda \; -->}$ is" - but isn't that the same thing?
Also: For the second case, is $t\ast <!--\; \ast \; -->e^{\mathrm{\lambda <!--\; \lambda \; -->}t}$ also a solution?

About differential equations.
1. While solving differential equations, is it okay to find an expression containing y and x (without y′) from which it's rather difficult to express y, for example,
$$\frac{{\ln }^2<!--\; \; -->(y)}{y}-<!--\; -\; -->\frac{x}{y}=C?$$
2. Does it make sense to try to find x(y) in some problems where it's easier?

Solving Coupled Differential Equations
I have the following differential equations, for modeling predator-prey relationships:
$$\frac{dx}{dt}=Ax-<!--\; -\; -->Bxy$$
$$\frac{dy}{dt}=Cxy-<!--\; -\; -->Dy$$
Where A, B, C, and D are constants. How could I go about solving this? I've only really worked with basic first order differential equations before, and I've found little to help me figure this out. Any help would be much appreciated.

I need help with Differential equations.
What is the Implicit answer for this differential equation?
$\frac{dy}{dx}=y^2-<!--\; -\; -->4$

Differential equations systems
We have that system of differential equations:
$$\{\begin{array}{l}{x}^{\prime }=-<!--\; -\; -->x\\ y^{\prime }=-<!--\; -\; -->2y\end{array}$$
I have to solve that system but I only know the method of derive first equation and substitute in the second and get a second order differential equation which I know to solve. But it seems that that method doesn't work here. How can I solve it? I'm beginner at this kind of exercises.

Distinction between "measure differential equations" and "differential equations in distributions"?
Is there a universally recognized term for ODEs considered in the sense of distributions used to describe impulsive/discontinuous processes? I noticed that some authors call such ODEs "measure differential equations" while others use the term "differential equations in distributions". But I don't see a major difference between them. Can anyone please make the point clear?

How are Linear Differential Equations formed?
A linear differential equation of second or higher order have more than 1 solutions of $y(y1,y2,$ etc.). Hence a common solution is obtained by assuming $y=c1y1+c2y2+.....$ Now my questions are:
1. How are linear differential equations formed from general algebraic expressions?
2. Why is it that a linear differential equation of second or higher order contain more than one solution of y that satisfy them?

Fourier and differential equations
Found this problem, just so you know it's my first time using Fourier to solve differential equations.
$$f^{″}(x)+f(x)=3\cos <!--\; \; -->(2x)$$

Determinants and Differential Equations Problem
"[Here] we explore a relationship between determinants and solutions to a differential equation. The $3\times <!--\; \times \; -->3$ matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations."
This is the question following the description:
"Verify that $y_1(x)=\cos <!--\; \; -->(2x),y_2(x)=\sin <!--\; \; -->(2x),y_3(x)=e^x$ are solutions to the differential equation: $y^{‴}-<!--\; -\; -->y^{″}+4y^{\prime }-<!--\; -\; -->4y=0$, and show that $\{\{y_1,y_2,y_3\},\{y_1^{\prime },y_2^{\prime },y_3^{\prime }\},\{y_1^{″},y_2^{″},y_3^{″}\}\}$ is nonzero on any interval."
Now I'm really just looking for an explanation on how I would show that it is nonzero on any interval. I completed the first part, I believe, by just taking up to the third derivative for each and plugging each set in to verify an identity of 0=0 at the end. I would then put the terms into the matrix which would give me:
$$\{\{\cos <!--\; \; -->(2x),\sin <!--\; \; -->(2x),e^x\},\{-<!--\; -\; -->2\sin <!--\; \; -->(2x),2\cos <!--\; \; -->(2x),e^x\},\{-<!--\; -\; -->4\cos <!--\; \; -->(2x),-<!--\; -\; -->4\sin <!--\; \; -->(2x),e^x\}\}$$
So I've gotten this far and if anyone could point me in the right direction on where to go from here it would be greatly appreciated.

Why aren't exact differential equations considered PDE?
Exact differential equations come from finding the total differential from some multivariable function.
In the exact differential equation $M\mathrm{d}x+N\mathrm{d}y=0$
M and N are considered to be partial derivatives of some potential function... So why aren't exact differential equations considered PDEs? After all, you're finding the potential function given it's partial derivatives...

Systems of Differential Equations and higher order Differential Equations.
I've seen how one can transform a higher order ordinary differential equation into a system of first-order differential equations, but I haven't been able to find the converse. Is it true that one can transform any system into a higher-order differential equation? If so, is there a general method to do so?