ODE: $$u''(t)+ \omega^{2} u(t)=0$$ I'd like to formulate the corresponding first-order linear ordinary differential equation. My attempt: Let $u(t) \neq 0$ $\forall t \in \mathbb{R} $, then $$u''(t)+\omega^{2} u(t)=0 \Leftrightarrow 2 \cdot u'(t) u''(t)+ 2 \cdot \omega^{2} u(t)\cdot u'(t)=((u'(t))^{2})'+\omega^{2} ((u(t))^{2})'=0$$ You can see: $$(\frac{u'(t)}{\omega \cdot u(t)})^{2}=1 \Rightarrow \frac{u'(t)}{\omega \cdot u(t)} = \pm 1 \Rightarrow u'(t)=\pm(\omega \cdot u(t))$$ But $u'(t)$ doesn't satisfy my ODE. Is there another way to get $u'(t)$?

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    $\begingroup$ I don't like that you write $u''(t)+\omega^{2} u(t)=2 \cdot u'(t) u''(t)+ 2 \cdot \omega^{2} u(t)\cdot u'(t)$ even if it's true because both sides are equal to $0$. It looks like you're the kind of person writing things like $2+3 = 5 + 8 = 13$. $\endgroup$ – md2perpe Jun 22 '17 at 21:23
  • $\begingroup$ How did you get $\left(\frac{u'(t)}{\omega\cdot u(t)}\right)^2 = 1$? $\endgroup$ – md2perpe Jun 22 '17 at 21:24

$((u')^2)' + \omega^2(u^2)' = 0$


$(u')^2 + \omega^2u^2 = C$


$(u')^2 = C - \omega^2u^2\\ \frac {u'}{\sqrt {C-\omega^2u^2}} =1$


$\frac 1{\omega}\arcsin \frac {u}{C} =t + \phi$


$u =C\sin(\omega t + \phi)$

However, I prefer this representation

$u =A\cos\omega t + B\sin\omega t$


I agree with you up to $ 0 = (u'^2 + \omega^2 u^2)' $. But we integrate this and find $$ u'^2 + \omega^2 u^2 = A^2\omega^2 $$ for some constant $A$ ($A^2$ being positive because the left-hand side is a sum of squares). This is the first-order equation. We note $$ \pm \omega (t-t_0) = \int_0^u \frac{du'}{\sqrt{A^2-u'^2}}, $$ and this integral has solution $\arcsin{(u/A)}$.


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