How to create a simple differential equation I am doing numerical analysis where we work with differential equations but I have never had any classes on differential equations. It seems you can get by in an introductory numerical analysis course with just knowing what a differential equation is an how the initial value problem solving process works. So I know a few process for solving them numerically but I have never learned how to solve them mathematically.
Anyway, I would like to know how to 'create' differential equation for testing my numerical approximation algorithms against, but as I have never learned how to deal with them mathematically I don't know the process for creating them. Can someone show me a step by step process for creating simple differential equations?
If I have $y = 5x^2 +c$, $c$ being some constant
I can make a differential equation of the form -
$\frac{dy}{dx} = 10x$
But I want to have differential equations where the $\frac{dy}{dx}$ is dependent on not just $x$, but on $x$ and $y$.
What is the process for doing this?
Edit
To clarify, I am looking to create simple equations of the form $\frac{dy}{dx} = f(x,y)$, $y(0) = y_0$ with solutions $y$ that are also relatively simple. This is so I can then write out the first few iterations of RK methods by hand to see how it all works.
 A: Consider the family of curves $\gamma_C$ given by an equation of the form
$$F(x,y)=C\ ,\qquad(1)$$
where the function $$F:\ {\mathbb R}^2\to {\mathbb R}\,\quad (x,y)\to F(x,y)$$
is explicitly given; e.g. $F(x,y):=x^2+y^2$. "Locally" such a $\gamma_C$ is the graph of a function $x\mapsto y(x)$. It follows that the function
$$x\mapsto\phi(x):=F\bigl(x, y(x)\bigr)$$
is identically equal to $C$. Therefore by the chain rule
$$\phi'(x)=F_{.1}\bigl(x,y(x)\bigr)\cdot 1+ F_{.2}\bigl(x,y(x)\bigr)\cdot y'(x)\equiv0\ ,$$
where $F_{.1}$ denotes the partial derivative of $F$ with respect to the first entry, and similarly for $F_{.2}$. Solving the last equation for $y'(x)$ gives
$$y'(x)\equiv -{F_{.1}\bigl(x,y(x)\bigr)\over F_{.2}\bigl(x,y(x)\bigr)}\ .$$
But this is saying that the function $x\mapsto y(x)$ satisfies the differential equation
$$y'=-{F_x(x,y)\over F_y(x,y)}\ .\qquad(2)$$
Here the right side ($=: f(x,y)$) is a known (resp., easily computable) function  of $x$ and $y$ which is independent of $C$. Therefore the equation $(2)$ can be considered as the differential equation characterizing the family of curves $\gamma_C$ defined by $(1)$.
Consider the following example: The set $\gamma_C$ of points $(x,y)$ whose distances to the "foci" $(\pm1,0)$ have given product $C>0$ is called a Cassini curve. The equation of this curve is given by
$$F(x,y):=\bigl((x-1)^2+y^2\bigr)\bigl((x+1)^2+y^2\bigr)=C^2\ .$$
When $C=1$ this curve is called a lemniscate. In order to obtain the differential equation of the family of curves $\gamma_C$ we compute
$$F_x=2(x-1)\bigl((x+1)^2+y^2\bigr)+2(x+1)\bigl((x-1)^2+y^2\bigr)=\ldots$$
and 
$$F_y=2y\bigl((x+1)^2+y^2\bigr)+2y\bigl((x-1)^2+y^2\bigr)=\ldots\quad.$$
After doing the calculations (and maybe some simplification) we can put down
the differential equation
$$y'=-{F_x(x,y)\over F_y(x,y)}=:f(x,y)\ ,$$
where now $f(x,y)$ is a certain definite expression in $x$ and $y$ not containing the parameter $C$.
Note that the above explanations do not take into account "singular points" or points where the tangent of the considered curve $\gamma_C$ is vertical.
A: I'm not entirely sure what you mean, but it sounds like you want to know how to differentiate implicitly. For instance, differentiating $x^{2}+y^{2}=25$ implicitly to get $\frac{dy}{dx}$ as a function of $x$ and $y$:
$$2x+2y\frac{dy}{dx}=0\implies \frac{dy}{dx}=-\frac{x}{y}$$
The process for doing this is by differentiating each term with respect to $x$ using the chain rule, for instance if we had $x^{2}+9xy^2+y^{4}=0$, then:
$$\frac{d}{dx}(x^{2})+\frac{d}{dx}(xy^{2})+\frac{d}{dx}(y^{4})=\frac{d}{dx}(0)$$
Therefore:
$$2x+y^{2}+2xy\frac{dy}{dx}+4y^{3}\frac{dy}{dx}=0 \implies (2xy+4y^{3})\frac{dy}{dx}=-2x-y^{2} \\ \therefore \frac{dy}{dx}=\frac{-2x-y^{2}}{2xy+4y^{3}}$$
I hope this answers your question?
A: Here is an example of "making" a differential equation you may be interested in solving using numerical methods. 
$$y = e^{-x^2}$$
$$y' = -2x e^{-x^2}$$
$$y' = -2x \left(e^{-x^2}\right)$$
$$y' = -2x \left(y\right)$$
$$y' = -2xy$$
The bottom equation would be a typical first order ODE that you would want to solve using something like Euler's Method or a Runge Kutta.
If you noticed it was actually quite easy to set this up. I took the derivative of something I knew and then found the original function in the derivative. Then I just replace that with a $y$ and pretend that I don't know the answer for $y$.
A: Usually the process to create differential equations from simple equations is by taking derivative of a variable based on the other ones, but speaking of how in the real world differential equations are created is like this:
Consider a society in which the rate of change of population to time (dp/dt) equals square of population (this is just an example in which "rate of change" depends on another variable), so we have :
dp/dt = p^2
This is now a first order non linear differential equation and you have a real world problem here exhibited in its mathematical form. 
So solving this differential equation gives you all possible functions which show the amount of population at any given time: i.e, P= f(t)
Now using simple integration we get to this solution:
p = 1/(c - t)
In which c is all possible constants that yeild to our unique differential equation.
I hope you have now a clear idea of where differential equations come from.
