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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.2012-06-26
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    If you take the minus sign, do you get $x'(0)=2$?2012-06-26
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    @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...2012-06-26
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    @BrianM.Scott - I don't know $x(t)$, so I can't answer your question2012-06-26
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    @Belgi Perhaps more explanation of how you solved the problem (and in particular, how you solved it without deriving an expression for $x(t)$ would help here?2012-06-26
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    I didn’t ask about $x(t)$.2012-06-26
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    What claim? You put your initial conditions to the test. If the plus sign works, great. If the minus sign also works, great. If one of them doesn't work, then you only have one solution to work with.2012-06-26
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    @Belgi You might argue that $x$ is continuous and doesn't ever take the value $0$.2012-06-26
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    I used the energy equation to find the values...I can't tell about $x'(0)$ if I pick the $-$ sign. I think @Cocopuffs argument is good since $x(0)>0$ and it is continuous and doesn't ever take the value 02012-06-26

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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 $u'(\pm t_1-\tau)=\pm\sqrt{2\ln3+4\,}$.