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I have a question in regard to the following;

considering the ODE

$y^{(4)}+2y''+y=0$

we can factor and find that the roots are $$r_1=r_2=i$$ and $$r_3=r_4=-i$$

So, I thought that a solution will be of the form $y(t)=c_1\cos(t)+c_2\sin(t)+c_3t\cos(-t)+c_4 t\sin (-t)$

but the answer given is the same except for it does not include the -t in the later two cos and sin.

I know that $cos(-t)=cos(t)$ but not for sin, so what is it I am doing wrong?

Thanks

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  • $\begingroup$ $c_4\sin(-t) = (-c_4)\sin(t)$. (But shouldn't there be a factor of $t$ in that term too?) $\endgroup$ – Henning Makholm May 18 '15 at 19:50
  • $\begingroup$ the negative sign is absorbed by the constant $c_4$. $\endgroup$ – Emilio Novati May 18 '15 at 19:51
  • $\begingroup$ Oh okay thanks, that makes more sense now. Lets say you did not do such, the answer would still be correct , no? $\endgroup$ – Quality May 18 '15 at 19:51
  • $\begingroup$ No. As noted by @Henning the solution has a term $ct\sin t$. $\endgroup$ – Emilio Novati May 18 '15 at 19:57
  • $\begingroup$ Oops yea I meant to have that when I wrote it up as well, other than that $\endgroup$ – Quality May 18 '15 at 19:59
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A fundamental set of solutions will be $$\{e^{it},e^{-it},te^{it},te^{-it}\}$$ (i.e. the four functions with the given formulas) as seen by direct examination of the roots of the characteristic equation. Taking linear combinations, (e.g. $\sin(t) = \frac{1}{2i}(e^{it}-e^{-it})$ ), we obtain another set of solutions: $$\{\sin t, \cos t, t\sin t, t\cos t\}$$ which is linearly independent, so the general solution can be expressed as $$y(t) = c_1\sin t + c_2\cos t + c_3 t\sin t + c_4t \cos t\, \quad c_i\in\Bbb C$$ or just $c_i\in \Bbb R$ for real-valued solutions.

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