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:
--- = 0
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
--- = k
This equation has as its solution F = kx + c. Again, you can plug this in and try it out.
--- = F
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:
---- = k2F
---- = -k2F
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 ).