Question

A student asked me something, and I'd like to post it here for kicks.\[\frac{1}{2}\frac{2}{(s-3)^2-4}=\frac{1}{s^2-6s+5}=\frac{1}{(s-5)(s-1)}=\frac{1}{4}\frac{1}{s-5}-\frac{1}{4}\frac{1}{s-1}\]We've shown that the very left of this expression is equal to the very right. Would this then imply that\[\frac{1}{2}\mathcal{L^{-1}}\left \{ \frac{2}{(s-3)^2-4} \right \}=\frac{1}{4}\mathcal{L^{-1}}\left \{ \frac{1}{s-5} \right \}-\frac{1}{4}\mathcal{L^{-1}}\left \{ \frac{1}{s-1} \right \}\]\[\implies \frac{1}{2}e^{-3t}\sin(2t)=\frac{1}{4}e^{5t}-\frac{1}{4}e^{t}?\]

Answer 1

I'm tempted to say yes, but interesting observation nonetheless.

Answer 2

I guess this is what's driving them (and us) crazy since there are "many different answers" for any given ODE.

Answer 3

Wait a minute, that should be the hyperbolic sine:
\[\frac{1}{2}e^{3t}\sinh(2t)=\frac{1}{4}e^{5t}-\frac{1}{4}e^t\]

Answer 4

\[\frac{1}{2}e^{3t}\left ( \frac{e^{2t}-e^{-2t}}{2} \right )=\frac{1}{4}e^{5t}-\frac{1}{4}e^t\]Duh.