Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Your argument should involve your initial conditions. –  sidht 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
1  
@Belgi You might argue that $x$ is continuous and doesn't ever take the value $0$. –  Cocopuffs Jun 26 '12 at 20:46
show 4 more comments

1 Answer

up vote 3 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.