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I want to solve $$ \ddot z + Az=0 $$ with $A>0$

I assume that $$z=e^{\lambda t} $$ So the equation becomes $$z(\lambda^2+A)=0$$ and so $\lambda_1=i\sqrt A$ and $\lambda_2=-i\sqrt A$ and then $$z=e^{i\sqrt At}+e^{-i\sqrt A t}=2\cos(\sqrt A t)$$

But I'm quite sure I'm actually supposed to find something looking like $$\alpha\cos(\sqrt At)+\beta\sin(\sqrt At)$$

What have I missed ?

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  • $\begingroup$ Is $A$ a positive number, or a positive definite matrix? $\endgroup$ – NoseKnowsAll Sep 17 '15 at 21:03
  • $\begingroup$ A positive number (this actually is a mechanics problem, $A=k/m$) $\endgroup$ – dcholleton Sep 17 '15 at 21:03
  • $\begingroup$ I don't see the problem, your answer is of the right form, it's just that $\beta=0$ $\endgroup$ – konewka Sep 17 '15 at 21:06
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    $\begingroup$ Setting $\lambda_i$ as given satisfies the equation, but your assumption that $z=e^{\lambda t}$ is overly restrictive. Assume instead that $z=k e^{\lambda t}$. $\endgroup$ – Alex Meiburg Sep 17 '15 at 21:07
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You have the correct eigenvalues, but you need to remember that a solution can be a linear combination of solutions:

$z = \eta e^{i \sqrt{A} t} + \gamma e^{-i \sqrt{A} t} $

Using Euler's Formula:

$z = \eta (cos(\sqrt{A}t)+sin(\sqrt{A}t)) + \gamma (cos(\sqrt{A}t)+sin(\sqrt{A}t)) $

then $\alpha$ and $\beta$ are some combo of $\eta$ and $\gamma$

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  • $\begingroup$ Thanks, I suppose that $\eta$ and $\gamma \in \mathbb{C}$ here ? $\endgroup$ – dcholleton Sep 17 '15 at 21:25
  • $\begingroup$ At the very least, yes, but I assume that they are in $\mathbb{R}$ $\endgroup$ – costrom Sep 17 '15 at 21:30

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