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 d^{2}s/dt^{2}). 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 = x_{0}e^{kt}, 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^{-1}y' = 3x. Divide both sides by x^{-1}, and then integrate to get: y = x^{3}. 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...

(x^{2} + y^{2})dx = -2x^{2}dy. As it sits, this is not separable. However, (a^{2}x^{2} + a^{2}y^{2})dx = a^{2}x^{2}dy is equivalent to a^{2}(x^{2} + y^{2})dx = a^{2}x^{2}dy, 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...

(x^{2} + v^{2}x^{2})dx = -2x^{2}(xdv + vdx)
x^{2}(1 + v^{2})dx = -2x^{3}dv + -2vx^{2}dx
x^{2}(v^{2} + 2v + 1)dx = -2x^{3}dv
x^{-1}dx = -2(v^{2} + 1)^{-2}dv.

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 M_{y} = N_{x}, 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 = e^{integral(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 = y^{n}R(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^{-n}y' + b(x)y^{1-n} = R(x)
z = y^{1-n}
dz = (1-n)y^{-n}dy
(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 = y_{c} + y_{p}

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 = e^{x} (using differential operator notation. D = d/dx)
y_{p} = Ae^{x} + Bxe^{x}
I'm dropping the 'p' so I can type less.
y' = Ae^{x} + Bxe^{x} + Be^{x}
Plug into the original equation...
Ae^{x} + Bxe^{x} + Be^{x} - Ae^{x} + Bxe^{x} = e^{x}
Be^{x} = e^{x} => B = 1
y_{p} = xe^{x}
y = y_{c} + y_{p}
y = c_{1}e^{x} + xe^{x}