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So we have one root, *q*. Let us say that we have two roots and the second one is close to the first. So *p*=*q*+*ε* is another root where* ε* is a small number. We have then

as our solution. We then have that

where the … represent things that are varying at least as *ε*^{2}*x*^{2}. We can get that from the series expansion of *e ^{x}* or reasoning around linear approximations of functions. So for small values of

*ε*we get that

Now let *C*+*D* be our new *D* and *Cε* be our new *C*. Remember that our old *C* can be chosen arbitrarily large as *ε* gets smaller and smaller, so the new *C* can in the end have any value. This gives us, as *ε* goes toward 0

as our general solution. The above is in itself not a proof, but we can make it a proof (at least that the above represents solutions) by substituting it into our original equation. We have that

and

This gives us

or, removing the common factor *e ^{qx}*,

This can be written as

The two first terms must be 0 since *q* is indeed the root of the characteristic equation. Also, since it is a double root we have that

So *a*=–2*q* and therefore is the last term zero too, and the suggested solution is a solution indeed.

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