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Is there any way to solve it without numerical way??

$$ \frac{d^2 y}{d x^2}= \frac{1}{y}$$ thanks in advance!!

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    Analytical Solution does exist : http://www.wolframalpha.com/input/?i=y%27%27%3D1%2Fy2012-10-19
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    Why don't you multiply by $y'$ in both sides and integrate? $$\mbox{Hint:}\quad y' y'' = \frac{1}{2} \big(y'^2\big)'$$2012-10-19
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    Got something from an answer below?2013-05-14

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Note that $y''y'=y'/y$ hence $(y')^2=c+2\log|y|$ hence $y'=\pm\sqrt{c+2\log|y|}$ and $$ \int_{y(0)}^{y(x)}\frac{\mathrm dt}{\sqrt{c+2\log|t|}}=\pm x. $$ The LHS does not seem to be (the inverse of) a usual function of $y(0)$ and $y(x)$. An equivalent formulation is $$ \mathrm e^{-c/2}\int_{\sqrt{c+2\log|y(0)|}}^{\sqrt{c+2\log|y(x)|}}\mathrm e^{t^2/2}\mathrm dt=\pm x, $$ and the LHS can be rewritten using the imaginary error function $\mathrm{erfi}$, with no obvious gain.

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    $x$ can be written in terms of $y$ using an erf function.2012-10-19
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    Tricky way (+1)2012-10-19
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    @BabakSorouh Classic trick in physics. Used in energy conserved systems. When a particle is under the action of a potential $V(x)$, where $x = x(t)$ is the position, the force is $f(x) = - \frac{d V}{d x}$, and the [equations of motion](http://en.wikipedia.org/wiki/Equations_of_motion) are $$m \ddot{x} = - \frac{d V}{dx}.$$ Multiplying by $\dot{x}$ both sides and integrating, one obtains $$E = \frac{1}{2} m \dot{x}^2 + V(x).$$ The first term is the [Kinetic Energy](http://en.wikipedia.org/wiki/Kinetic_energy) and the second the [Potential Energy](http://en.wikipedia.org/wiki/Potential_energy).2012-10-19
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    @BabakSorouh Thanks.2012-10-19