A differential equation, is an equation where there exist variables for a function f: X -> Y ( X,Y Banach spaces), and its derivates, both ordinary and partial allowed.

Let X,Y be Banach spaces, Z the set of functions X -> Y, D the set of derivates, both partial and ordinary allowed, which do not have to be defined on the whole X.
An (implicit) differential equation is an equation
F(f,x,df1,...,dfn) = 0
where F is a function F: Z x X x Y x D x ... x D -> Ym, the D-parameters dfi must be exactly same type of derivative for a fixed i, m > 0 an integer.
A function f: X -> Y is called a solution iff F applied to f and its derivatives needed for F is 0 for an non-empty (open ?) subset of X x Y.

This definition is quite ugly, but should capture all types of differential equations like ordinary differential equations or partial differential equations.

Examples:

• f(t) = f'(t) equivalent to f(t) - f'(t) = 0, the first form is called an explicit differential equation, this form doesn't always have to exist.
• ```df        df
--(t,x) = --(t,x)
dt        dx
```
• F'(A)H = - H-1AH , where A,H invertible matrices, diff. eqn. on the space of matrices.
Solution: F(A) = A-1

### What the heck is it!?

Relax, it's not as confusing as all that. A differential equation looks a lot like the equations you encounter in plain old algebra -- 2y-4x = 5 and the like. The word "differential" means that one of the variables (x's and y's to you non-mathers) in your equation is a derivative of a function.

If you haven't studied The Calculus -- and who can blame you, really? -- a derivative is a special function derived from another function. If you graph a function on a grid (or in space), and for some reason you want to know what the tangent line (or plane) is at every point of that function, you would compute that function's derivative. This tangent tells you how quickly your function is changing, and in what direction.

Sounds boring, I know, but it's a lot more useful than it sounds. For instance, if you have a function (y) telling you what the position of an object is, the derivative of that function (y' -- say "why-prime") would tell you its speed (how fast it's changing position), and the derivative of that derivative (y'' -- say "why-double-prime") would tell you its acceleration (how fast it's changing speed).

Now, let's go back to our algebra equation: 2y-4x = 5. If you wanted to know what y was, in terms of x, you'd shuffle things around according to what your math teacher told you and come up with y = 2x+5/2. But if we replaced y with its own derivative, and handed you 2y'-4x = 5, you'd be stuck, because you don't know calculus. But if you did, you'd solve for y' -- y' = 2x+5/2 -- and integrate it. That's like finding the derivative, but backwards. The solution happens to be y = x2+5x/2+c, where c is any real number you like. Yes, any real number -- this means that, technically, there are an infinite number of y's for any one value of x. Professional mathematicians have been trying for centuries to get rid of this inconvenience and make their problem-solving much easier, but it looks like we're stuck with it. If you're lucky, you're in a situation where you can set c = 0 and lock things down, but don't bet on it.

That wasn't a differential equation, though. That was just calculus. Differential equations are when you have both y and y' in the same equation -- 2y'-4xy+x = 5, for instance. You can't solve for y with a simple integral this time, but you can solve for it. You'll still have that nasty, inconvenient c in your solution, but if you have what's called an initial condition -- where y equals some number when x is zero -- you can get rid of it.

If your equation has just y and y' in it, it's called a first order differential equation. If you have y'' in it, it's a second order equation. A linear differential equation -- which means y and all its derivatives are multiplied by x's and numbers instead of each other -- is easier to solve than a nonlinear one.

### How on earth do I solve one?

I'm not going into detail on this, partly because it would require an entire textbook but mostly because I nearly failed the class myself. The reason I nearly failed the class is because there is no way to solve a differential equation.

Well, that's not exactly true. There's no one, universal method for solving all differential equations. Worse yet, the answer you get may look entirely different from the answer someone else gets, and both of you will still be correct. This makes it very hard to check your work in the back of the textbook, among other things.

However, if your equation happens to be one of a number of special types, then there are tried and true ways to solve them. The hard part is recognizing an equation as a certain type and remembering which types are solved using what techniques.

It's not fun, especially if you hated calculus. Here's a quick step-by-step on how to solve a first order linear differential equation like y'+tan(x)y = cos2(x), which I chose because it's already been solved right here next to me. A first order linear differential equation can always be written in the form y'+f(x)y = g(x), where f(x) and g(x) are some two functions of x, and only x.

First, find something called the integrating factor, which is a fancy term meaning "special number which makes it easier to solve this problem." The integrating factor for these types of equations is e to the power of the integral of whatever you're multiplying by y (in this case, tan(x)). After much flipping through your calculus notes, you're able to simplify this integrating factor to sec(x).

Second, multiply the integrating factor by whatever isn't multiplied by y or y' -- in this case, cos2(x) -- and integrate that. Again, your calculus notes eventually tell you that the integral of sec(x)cos2(x) is sin(x)+c (there's that nasty c again).

Third, divide what you got in the second step by the integrating factor to get y. In our example, we now have y = (sin(x)+c)/sec(x) = sin(x)cos(x)+c*cos(x). If you have an initial condition like, oh, y(0) = 2, you can solve for c: sin(0)cos(0)+c*cos(0) = 2 and, after a little more time with your notes, c=2. Our final answer is y = sin(x)cos(x)+2cos(x), and you can breathe again.

If a differential equation isn't one that can be solved neatly, there are still ways to break it down to a not-so-neat but useable solution. LaPlace transforms and Fourier transforms are the usual ways to go about this.

### Why would I want to do this?

You'd be surprised. You probably won't want it in everyday life, of course, but scientists of all sorts run into equations like this in their daily work. Radioactive matter (like plutonium or carbon-14) decays at a rate which changes according to how much hasn't yet decayed, and this can be represented by a linear differential equation. Similarly, the rate of change of an area's population will change depending on how many people are there. Newton's law of cooling is a linear differential equation, too, and can be used to determine how long a body has been dead in a room (along with other, less gory applications).

And then there are the more advanced, not-for-the-faint-of-heart applications that engineers and physicists encounter, like multiresolution analysis, fuzzy logic, electromagnetic fields, cellular growth in an organism, and the engineering of structures that are subject to variable stress like bridges and skyscrapers.

In short, whenever you need to model something that changes depending on how much it's changed, you'll need a differential equation. This actually includes most of the stuff in the universe. With a few special exceptions.

### I don't need any of that. I just want to balance my checkbook every month.

Fine with us. Interested in knowing what it took to design the pocket calculator you're using, though?... Hey, come back here!...

This is why you took three courses of calculus.

If you think about differential equations, all they really do is describe the rate of change of something. As mblase suggested, you could use differential equations to figure out velocity (change in position set against change in time: ds/dt) and acceleration (the change in the change in position set against change in time set against change in time: dV/dt or d2s/dt2). Growth problems work in a similar manner, where you start out with a simple equation like

dx/dt = kx

In this case, x is the population, and k is some constant of proportionality. The equation is fairly self-explanatory; you're measuring the change in population against the change in time, and it's equal to the population times some constant. After separation of variables, integration, and solving for C, you'll get something that looks like x = x0ekt, which you can use to find populations of bacteria cultures at any time. That's the key to most differential equations. You might get lucky and find out that the number of bacteria doubles every t hours and time your growths just right, but why do that when you can get a spiffy little equation that will tell you exactly how many critters you have at a specific time? The same applies to finding out continuous interest rates, radioactive decay, and any other equation of the sort.

Of course, there are my instructor's favorite, brine problems. Basically you deal with a tank of water with or without some quantifiable amount of evenly-disperesed impurity (i.e. salt) in it, then pump water (with or without impurities) into the tank and pump water out of the tank. Basically, you need information pertaining to the amount of water and impurity in the tank at time t=0, the rates at which the water is being pumped into and out of the tank, and the rate at which the impurity is being pumped into the tank. The general concept is that you have some function, call it Q'(t) (also, dQ/dt), that is defined as the rate the impurity is being added to the tank - the rate at which the impurity is being pumped out of the tank. Now, since we have a dQ/dt, we must have a Q(t), and we'll consider that to be the amount of impurity in the tank at any time t. I (won't) spare you the details, but generally the rate at which impurity is added will be fairly simple (so many pounds per gallon of impurity are in the water being added, and the water is being pumped in at so many gallons per minute. Multiply through for lbs/min, and that'll give you the rate of addition to the tank), and the rate at which the impurity is being taken out is some unholy figure (Q(t)/(original amount of water in the tank + (rate water is entering - rate water is leaving)*t) will give you lbs/gal, so multiply through by the rate the water is being pumped out (gal/min) to give you lbs/min). You'll end up with a linear differential equation which can be solved fairly easily, and from there, you can figure out how many pounds (or any unit of weight, really) of impurity you have, what concentration said impurity is, and the theoretical amount of solute the tank can hold (as time approaches infinity). All in all, it's pretty nifty stuff.

Then there are methods of solving ordinary differential equations...

Separation of Variables

This is fairly straight-forward. If you can separate the variables of an equation, usually by factoring out and dividing, then integrating both sides of the equation. For example...

x-1y' = 3x. Divide both sides by x-1, and then integrate to get: y = x3. Again, fairly straightforward.

Homogeneous Equations

In this sense, a function is said to be homogenous if f(ax,ay) = af(x,y). If an equation is homogeneous, it can be made separable through substitution, either x = vy or y = vx, whichever works best (this is not always apparent). For example...

(x2 + y2)dx = -2x2dy. As it sits, this is not separable. However, (a2x2 + a2y2)dx = a2x2dy is equivalent to a2(x2 + y2)dx = a2x2dy, and the 'a' term can cancel out. Thus, f(ax,ay) = af(x,y) and the equation is homogeneous. For simplicity, let's say y = vx. Then, dy = xdv + vdx. Substitute in...

```(x2 + v2x2)dx = -2x2(xdv + vdx)
x2(1 + v2)dx = -2x3dv + -2vx2dx
x2(v2 + 2v + 1)dx = -2x3dv
x-1dx = -2(v2 + 1)-2dv.
```

Now you simply integrate with whatever you have handy (pencil and paper, TI-89, Derive, etc.) and arrive at an answer (don't forget your c!). Since you want the solution in terms of x and y, just remember than y = vx, so v = y/x. Substitute in and the equation is solved.

Exact Equations

Ah! Yet another method! Equations are said to be exact of My = Nx, with an equation in the form of Mdx + Ndy = 0, and the subscripts mean the partial derivitive. If they do match, simply integrate Mdx and Ndy, and take a psuedo-union of their integrals. Basically, if, say, xy results from integrating both, then simply include it once in the final answer. This union is set equal to c. Note, there is a more complicated method than this that is essential for some advanced concepts, but for calculation, this will always work.

Linear Equations

An equation is said to be linear if it is of the form:

`y' + b(x)y = R(x), where b(x) and R(x) are functions of x alone.`

The key to this type of equation is to find the integrating factor, which is simply v = eintegral(b(x)dx)). Then, multiply the original equation by this factor to get v(y' + b(x)y) = vR(x). Integrate both sides, and you will get yv = integral(vR(x)dx). Divide by v, and your equation is solved. (notice that integrating the LHS will result in the single y term being multiplied by the integrating factor).

Bernoulli Equation

A Bernoulli equation is of the form.:

`y' + b(x)y = ynR(x)`

Ah! This looks very similar to our linear equation, but it has that pesky y term on the RHS. Again, substitution will save us...

```y-ny' + b(x)y1-n = R(x)
z = y1-n
dz = (1-n)y-ndy
(1-n)*dz/dx + b(x)z = R(x)
z' + (b(x)/(1-n))z = (1-n)R(x)```

And it is easy to see that the equation is linear with respect to z. Use the method for linear equations, and then substitute back in for z.

Auxiliary Equation

See the node.

Method of Undetermined Coefficients

This method works well for non-homogeneous equations where you have a vague idea from where the RHS came. The general equation will be of the form:

`y = yc + yp`
where the former is the complementary equation and the latter is a particular solution to the equation.

The complementary equation is simply the auxiliary equation of the LHS = 0. The particular equation is arrived at by equating coefficients, a method that I will describe below...

```(D - 1)y = ex (using differential operator notation. D = d/dx)
yp = Aex + Bxex
I'm dropping the 'p' so I can type less.
y' = Aex + Bxex + Bex
Plug into the original equation...
Aex + Bxex + Bex - Aex + Bxex = ex
Bex = ex => B = 1
yp = xex
y = yc + yp
y = c1ex + xex```

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