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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 : wolframalpha.com/input/?i=y%27%27%3D1%2Fy –  Inquest Oct 19 '12 at 17:54
Why don't you multiply by $y'$ in both sides and integrate? $$\mbox{Hint:}\quad y' y'' = \frac{1}{2} \big(y'^2\big)'$$ –  Pragabhava Oct 19 '12 at 17:58
Got something from an answer below? –  Did May 14 '13 at 7:24

1 Answer 1

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. –  GEdgar Oct 19 '12 at 18:24
Tricky way (+1) –  Babak S. Oct 19 '12 at 18:26
@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 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 and the second the Potential Energy. –  Pragabhava Oct 19 '12 at 19:05
@BabakSorouh Thanks. –  Did Oct 19 '12 at 20:23

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