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Given $\lambda = a+iw$ and $\lambda = a-iw$

I then have the 2 real solutions given by: $$y(t) = Re(e^{\lambda t}) = e^{at}\cos(wt)$$ and $$y(t)=Im(e^{\lambda t}) = e^{at}\sin(wt)$$

I then have to show that these $2$ solutions are linearly independent - I tried to write the equation:

$$c_1e^{at}\cos(wt)+c_2e^{at}\sin(wt)=0$$

And then I want to show that the only solution to this equation is that $c_1=c_2=0$, but I can't really figure out how to show it. I usually just differentiate the function so that I have 2 equations with 2 unknowns, but I didn't really get anything good out of trying to solve that :)

N. F. Taussig
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some_name
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    Please see this tutorial on how to typeset mathematics on this site. To produce a Greek letter such as $\lambda$, type \lambda in math mode. To produce $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, $\cot x$, type \sin x, \cos x, \tan x, \csc x, \sec x, and \cot x, respectively, in math mode. – N. F. Taussig Sep 26 '15 at 12:12
  • ooh, okay - I'll remember that :) – some_name Sep 26 '15 at 12:19

1 Answers1

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Take two values of $t$. One for which the sinus vanishes but not the cosinus and vice versa for the second one.

mathcounterexamples.net
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