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Why is the solution to the following differential equation: $$0=-x^2 \left({dt\over d\mu}\right)^2+\left({dx\over d\mu}\right)^2----(*)$$ where $x(\mu)\Big|_{\mu=0}=x_0$ and ${dx\over d\mu}\Big|_{\mu=0}=v_0$

equal to $$x(\mu)=\sqrt{2v_0x_0\mu+x_0^2}$$?

I got a different solution: Initial conditions give $$\left({dt\over d\mu}\right)^2={v_0^2\over x_0^2}$$ $$(*)\implies {dx\over d\mu}=\pm {v_0\over x_0} x$$ $$\implies x(\mu)=c\exp(\pm{v_0\over x_0}\mu)$$

What went wrong?

Context : The equation is derived as a null geodesic equation $\left({d s\over d\mu}\right)^2 = 0$ for the metric $$ds^2=-x^2dt^2+dx^2$$. So I believe $\mu =at+b$ for some constants $a,b$. Hence justifying the setting $$\left({dt\over d\mu}\right)^2$$ to be a constant.

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  1. (*) is one equation with two unknown functions $t,x$. Generally, you need as many equations as unknown functions, in order to have a unique solution.
  2. Apparently, you work around the issue 1. by assuming that $dt/d\mu$ is constant. Then indeed, your formula $dx/d\mu $ is a constant multiple of $x$ and the solution is exponential.
  3. Whoever derived the solution $x(\mu)=\sqrt{2v_0x_0\mu+x_0^2}$ may have been using a different function $t(\mu)$.
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Right. It's hard to see how you can start with an equation with $t$ in it and get an answer without. – Gerry Myerson Feb 24 '13 at 3:31

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