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Like integrals, differential equations are solved mostly by guessing. Unfortunately, there is little in the way of hard and fast rules. When we're solving a differential equation, we're looking for some function, F, that fits some particular prescription. The simpliest differential equation is:

d F

--- = 0

d x

This differential equation has as its only solution F(x) = constant. You can see this solution works by plugging the function into the equation and checking.

This differential equation is called a 1st order equation - there's only 1st order derivatives. Like polynomials, we're looking for as many solutions as there are orders. When we find that many solutions, we're done: the complete solution to the differential equation is the weighted sum of the seperate solutions.

The next simpliest differential equation is

d F

--- = k

d x

This equation has as its solution F = kx + c. Again, you can plug this in and try it out.

Next,

d F

--- = F

d x

That is, the derivative of the function is the function. We recognize
the answer to this, F = e^{x}. There's no rule that makes this
obvious; we just remember, in exactly the same way we remember things when
we do a crossword puzzle.

A very important pair of differential equations are:

d^{2}F
---- = k ^{2}F
dx ^{2} |
d^{2}F
---- = -k ^{2}F
dx ^{2} |

These second-order equations come up in physics all the time. It's important
to recognize them instantly and see the various solutions instantly. The
solution to the first equation is F = ae^{kx} + be^{-kx}.
These solutions have the rather unfortunate property that they're neither
manifestly symmetric nor anti-symmetric. This can be easily fixed, however.
Any function F(x) can be resolved into a symmetric and an anti-symmetric
part,

F_{symmetric} = ( F(x) + F(-x) ) / 2

F_{anti-symmetric} = ( F(x) - F(-x) ) / 2

F = F_{symmetric} + F_{anti-symmetric}

We can play the same trick with the solutions to the differential equation above,

F_{symmetric} = (e^{kx} + e^{-kx}) / 2 = cosh(
kx )

F_{anti-symmetric} = = (e^{kx} - e^{-kx})
/ 2 = sinh( kx )

Now our solutions will be F = a cosh( kx ) + b sinh( kx ).

The solution to the second equation is F = ae^{ikx} +
be^{-ikx}. Again, these solutions are neither manifestly symmetric
nor anti-symmetric. We can fix this with the same trick,

F_{symmetric} = (e^{ikx} + e^{-ikx}) / 2 = cos(
kx )

F_{anti-symmetric} = = (e^{ikx} - e^{-ikx})
/ 2i = sin( kx )

Now our solutions will be F = a cos( kx ) + b sin( kx ).