# Deciding on what solution, $x''=\frac{1}{x}$

I am trying to solve the following:

Let $x(t)$ be a solution of $x''=\frac{1}{x}$ satisfying $x(0)=1,x'(0)=2$. Let $t_0$ be the time when $x(t_0)=3$, find $x'(t_0)$.

I got that $x'(t_0)=\pm\sqrt{4+2ln(3)}$.

I believe I should take the $+$ sign solution, how can I argue that ? Help is appriciated!

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Your argument should involve your initial conditions. –  jip Jun 26 '12 at 20:35
If you take the minus sign, do you get $x'(0)=2$? –  Brian M. Scott Jun 26 '12 at 20:35
@jak 0 I understand that the initial condition got something to do with this, but I don't have an argument to prove my claim... –  Belgi Jun 26 '12 at 20:35
@BrianM.Scott - I don't know $x(t)$, so I can't answer your question –  Belgi Jun 26 '12 at 20:39
@Belgi You might argue that $x$ is continuous and doesn't ever take the value $0$. –  Cocopuffs Jun 26 '12 at 20:46

Both are valid. The $+$ corresponds to a value $t_0>0$; the $-$ to a value $t_0<0$. Multipliying the equation by $x'$ and integrating we obtain the energy equation $$(x')^2=2\ln x+4.$$ Since the equation is autonomous (the variable $t$ does not appear explicitely) solutions are invariant under translations: if $u(t)$ is a solution, so is $u(t+\tau)$ for all $\tau\in\mathbb{R}$. Consider the unique solution $v(t)$ such that $v(0)=e^{-2}$ and $v'(0)=0$. $v$ satisfies the same energy equation as $u$, so that there is a $\tau\in\mathbb{R}$ such that $u(t)=v(t+\tau)$. Moreover $v$ is even. There exists $t_1>0$ such that $v(\pm t_1)=3$, $v'(\pm t_1)=\pm\sqrt{2\ln3+4\,}$. Then $u'(\pm t_1-\tau)=\pm\sqrt{2\ln3+4\,}$.