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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

3 Answers 3

up vote 1 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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Wolfram alpha gives this result which shows it is not easy to find.

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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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