# How to solve $\ddot z + Az = 0$

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 ?

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

$z = \eta e^{i \sqrt{A} t} + \gamma e^{-i \sqrt{A} t}$
$z = \eta (cos(\sqrt{A}t)+sin(\sqrt{A}t)) + \gamma (cos(\sqrt{A}t)+sin(\sqrt{A}t))$
then $\alpha$ and $\beta$ are some combo of $\eta$ and $\gamma$
• Thanks, I suppose that $\eta$ and $\gamma \in \mathbb{C}$ here ? – dcholleton Sep 17 '15 at 21:25
• At the very least, yes, but I assume that they are in $\mathbb{R}$ – costrom Sep 17 '15 at 21:30