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A long time ago I was curious about the closed-form solutions to the equation:

$ \frac{d^{n}y}{dx^n} y = 1$.

For $n$ an odd number, try $y = A x^k$. Then $y^{(n)} = A k(k-1)...(k-n)x^{k-n}$. This gives the formula

$A^2 k(k-1)...(k-n) x^{2k - n - 1} = 1$

Which can only be true if $k = \frac{n+1}{2}$ and $n$ is odd ($k$ cannot be an integer for the formula to work, check this yourself). Furthermore one has to have that

$A = (k(k-1)...(k-n))^{-1/2}$

which is real if $n$ is odd.

Thus there are closed-form solutions to my problem for $n$ odd, and my question is if anyone can find a closed-form solution for $n=2$ or in general if $n$ is even.

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5  
Mathematica says: Inverse error function. wolframalpha.com/input/… –  KennyTM Aug 16 '10 at 15:31

2 Answers 2

up vote 6 down vote accepted

Let $y=y(x)$ and write primes for derivatives w.r.t $x$. In the equation $$y''y=1$$ the independent variable $x$ does not appear, so we can do a (somewhat magical) change of variables: in the new equation, the independent variable will be $y$ and the dependent variable will be $p=y'$. If we use dots to denote derivatives with respect to $y$, a little computation using the chain rule shows that $p''=p\dot p$, so the equation becomes $$\dot ppy=1.$$ This can be rewritten as $$p\,\mathrm dp=\frac{\mathrm dy}{y},$$ which has its variables separated. It can be integrated at once, to get the relation $$\frac{p^2}{2}=\log y+c$$ with $c$ a constant. Now we recall what $p$ was, that is $y'$, and this becomes a new (first order!) ODE on $y(x)$. We have «simplified» things... (We could have obtained the same reduction od order by multiplying the original equation by $y'$, dividing it by $y$ and integrating: the trick with the $p$ substitution is, though, a general method)

The same approach will similarly simplify the general case.

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Well I tried for $\frac{d^2y}{dx^2}y=1$, and I have partial solution. I will use this reply to trace the work since it supports LaTeX, I will try to complete it later (or someone else can take over).

Suppose $x=f(y)$, so $y=f^{-1}(x)$.

Then $dx=f'(y)dy$,

or $\frac{dy}{dx}=\frac{1}{f'(y)}$.

So $\frac{d^2y}{dx^2}=\frac{d}{dy}\left(\frac{1}{f'(y)}\right)\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^2}\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^3}=-\frac{g'(y)}{(g(y))^3}$ where $g(y)=f'(y)$.

Solving $-\frac{g'(y)}{(g(y))^3}=\frac{1}{y}$, we have

$g(y)=f'(y)=\frac{1}{\sqrt{\log(y^2)+c}}$.

Now we need to solve for $f(y)$, and invert the function to get $y=f^{-1}(x)$.

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