# Classical Electromagnetism

## Appendix 3: Differential Equations

### By Mark Lawrence

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## A Basic Introduction to Differential Equations

Differential equations come up all the time in physics. This appendix can be considered a review / primer. It's not a substitute for a proper course.

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 = ex. 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:

 d2F ---- = k2F dx2 d2F ---- = -k2F dx2

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 = aekx + 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,

Fsymmetric = ( F(x) + F(-x) ) / 2

Fanti-symmetric =  ( F(x) - F(-x) ) / 2

F = Fsymmetric + Fanti-symmetric

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

Fsymmetric = (ekx + e-kx) / 2 = cosh( kx )

Fanti-symmetric =  = (ekx - 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 = aeikx + be-ikx. Again, these solutions are neither manifestly symmetric nor anti-symmetric. We can fix this with the same trick,

Fsymmetric = (eikx + e-ikx) / 2 = cos( kx )

Fanti-symmetric =  = (eikx - e-ikx) / 2i = sin( kx )

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