Function for (damped) SHM as a linear combination of two exponentials

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quozzy
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So, in lectures we derived the equation for damped SHM by solving the differential equation relating position (x), mass (m), spring constant (s), and damping coefficient (r):

[tex]m\ddot{x}=-\frac{s}{m}x-r\dot{x}[/tex]

Using a solution of the form [tex]Ae^{\alpha t}[/tex], we find that:

[tex]x=Ae^{-pt}e^{\pm qt}[/tex],

where [tex]p=\frac{r}{2m}[/tex], and [tex]q=\sqrt{p^{2}-\frac{s}{m}}[/tex].

Everything until and including this I understand. However, the final step, with no explanation, turns the solution into the following:

[tex]e^{-pt}(C_{1}e^{qt}+C_{2}e^{-qt})[/tex],

where C1 and C2 are some arbitrary constants. (i.e. a linear combination of the two distinct solutions.) In trying to research this online, I found an article that mentions the solution holds true for all complex values of C1 and C2. I don't understand how, algebraically, you can go from the previous step to the last one. Somebody help me out?

Thanks in advance.

P.S. I don't know why the closing bracket doesn't show up in the last equation, but it should be there. (EDIT: Nevermind, it works now.)
 
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It is true that if f(x) and g(x) are distinct solutions of a linear equation then their sum is also a solution as is any arbitrary linear combination of them.

Note this is true of any linear equation, not only differential ones.

In your penultimate line you have two distinct functions obtained by taking either the = or the - sign in

[tex]{e^{ \pm qx}}[/tex]

Does this help?

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