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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
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Why don't you multiply by $y'$ in both sides and integrate? $$\mbox{Hint:}\quad y' y'' = \frac{1}{2} \big(y'^2\big)'$$ – Pragabhava Oct 19 '12 at 17:58
Got something from an answer below? – Did May 14 at 7:24

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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. – GEdgar Oct 19 '12 at 18:24
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Tricky way (+1) – Babak S. Oct 19 '12 at 18:26
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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 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
up vote -3 down vote

It is easier to solve the differential equation as $x$ is a function in $y$

$$\frac{d^2y}{dx^2} = \frac{1}{y} \implies \frac{d^2 x}{dy^2} = y \implies \frac{d x}{dy}=\frac{y^2}{2}+c_1 \implies x =\frac{y^3}{6}+c_1y+c_2 \,. $$

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That doesn't seem correct: $$\frac{d}{dx} x = 1 =\frac{d}{dx} \left(\frac{y^3}{6}+c_1 y + c_2\right) = \frac{y^2 y'}{2} + c_1 y'$$ $$ \frac{d}{dx} 1 = 0 = \frac{d}{dx} \left(\frac{y^2 y'}{2} + c_1 y'\right)= y y'^2 + \frac{y^2 y''}{2} + c_1 y''$$ – Pragabhava Oct 19 '12 at 18:40
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This is why: $$\frac{d y}{d x} = \frac{1}{x'(y)} \Longrightarrow \frac{d^2 y}{d x^2} = \frac{d}{dx} \frac{1}{x'(y)} = \frac{dy}{dx} \frac{d}{dy}\left(\frac{1}{x'(y)}\right) = - \frac{x''(y)}{x'(y)^2}$$ But as did point's out, the substitution can lead to a quadrature. – Pragabhava Oct 19 '12 at 18:46
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It is tempting to believe that $\frac{d^2 x}{dy^2}$ is the reciprocal of $\frac{d^2y}{dx^2}$. But one can check, for example using $y=x$ (or $y=x^3$) that this is not necessarily so. – André Nicolas Oct 19 '12 at 18:56
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@MhenniBenghorbal Nobody is questioning the technique, just that you are implementing it wrong. In that matter, I made a mistake also (:blush:), $$\frac{d^2 y}{d x^2} = - \frac{x''(y)}{x'(y)^3}.$$ If you don't believe me, do the math. – Pragabhava Oct 19 '12 at 19:19
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@Pragabhava Good luck--but be prepared for a loooong discussion. – Did Oct 19 '12 at 20:22
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