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The equation is:

$$\frac{d^2y}{dx^2} = \frac{100}{y}$$

Also if $y = f(x)$,

$f'(0) = 0$ and $f(0) = 10$

How would you solve this for $y$?

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The first idea that comes to my mind is to multiply both sides by $y'$ and integrate. But this leads to an order 1ODE which I can't solve at first sight. So this is probably a well-known type of ODE which I forgot... – 1015 Feb 15 '13 at 22:27
What are the initial conditions? – user7530 Feb 15 '13 at 22:41
up vote 3 down vote accepted

$$y''=\frac {100} y$$ is autonomous (ie $x$ is not present in the equation). The standard procedure therefore is to substitute $u(y)=y'$ with $$ y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u'u $$ solve $$ u'u = 100/y $$ with $u'u = \frac{1}{2}(u^2)'$ we get $$ u(y)^2 = 2\left(c_1 + 100\ln(y)\right). $$ This equation "can be integrated" $$ y' = \pm\sqrt{2\left(c_1 + 100\ln(y)\right)} $$ $$ \Rightarrow \int\frac{1}{\sqrt{2\left(c_1 + 100\ln(y)\right)}}dy = \mp\int 1\,dx $$ This is the point where it gets ugly...

The integral can be evaluated to $$ \sqrt{\frac{\pi}{200}} e^{-c_1/100}\operatorname{erfi}\left(\sqrt{\frac{c_1}{100}+\ln(y)}\right) + c_2 = \mp x $$ where $\operatorname{erfi}$ is the imaginary error function. Solve for $y$ and you are done...

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Multiply both sides by $y'$ and integrate. You will get here:

$$\frac{dy}{dx}=\pm\sqrt{2k_1+200\log (y)}$$ $$\int^x\frac{dy}{\sqrt{2k_1+200\log (y)}}=x+k_2$$

That is not an elementary integral, you can't get a solution as elementary functions. Wolfram uses the error function to give a solution that has no integrals in it.

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This answer is similar to that of the user example, but I though this answer might give you so more intuition about how to solve similar problems and when certain things are equal to each other.

In physics when you have a dynamic system where the force/acceleration on an object only depends on its position, then the potential energy of that object will also only be a function of its position and energy will be conserved. In this case the force $F$ can be expressed as follows

$$ y'' = F = -\nabla U(y) = \frac{100}{y}, $$

such that the potential $U(y)$ can be found with

$$ U(y) = -\int{\frac{100}{y}dy} = -100 \log{y} + c_1. $$

In physics the conservation of energy means that the sum of the kinetic energy and potential energy are constant

$$ \frac{y'^2}{2} - 100 \log{y} + c_1 = c_2, $$

$$ y' = \frac{dy}{dx} = \sqrt{200 \log{y} + c_3}. $$

By applying separation of variables the solution can be found with

$$ \int\frac{dy}{\sqrt{200 \log{y} + c_3}} = x + c_4, $$

which has the implicit solution

$$ \sqrt{\frac{\pi}{200}} e^{-\frac{c_3}{200}} \text{erfi}\left(\frac{c_3}{200} + \log(y)\right) = x + c_4, $$

where $\text{erfi}(x)$ is the imaginary error function ($\text{erfi}(x) = -i\, \text{erf}(i\, x)$). Rewriting this to an expression for $y(x)$ yields

$$ y(x) = \exp\left(\text{erfi}^{-1}\left(\sqrt{\frac{200}{\pi}} e^{\frac{c_3}{200}} (x + c_4)\right) - \frac{c_3}{200}\right). $$

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Wolfram alpha gives this result which shows it is not easy to find.

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