# $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$.

I'm looking at a simple differential equation of the form: $$y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right].\tag{\star}$$ Here $y=y(x)$ is a function of $x$ only, $y_0$ is a constant, and $x,y,y_0 \in \mathbb{R}$ are real.

After an hour or so working on a problem, Eq. $\left(\star\right)$ popped out. I haven't made a significant effort to solve this DE just yet, because it is so simple looking and I thought it'd be a nice question for this site. This is "work-work", not "homework".

Is there a name for this DE? Are there simple solutions, besides the trivial solution $y(x) = y_0$?

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Thanks to GEdgar and O.L., we have an implicit solution right away. Unfortunately, inverting that solution doesn't look like much fun! Now the search for explicit solutions begins... – Douglas B. Staple Jun 24 '13 at 14:11
For $y_0=0$ there is a solution $y(x)=-2/x^2$. – Start wearing purple Jun 24 '13 at 14:22
@O.L. Well, we can't have $y_0 = 0$ because of the $1/y_0$ in the log. I guess what you wrote had $y_0=0$ substituted on the LHS, and $y_0=1$ on the RHS. Tricked me too. Still interesting, though. – Douglas B. Staple Jun 24 '13 at 14:49
@DouglasBStaple We can actually. The expression for the 2nd derivative on the right does not contain $y_0$ at all since $\ln y/y_0=\ln y-\ln y_0$. I don't think that for $y_0\neq0$ one can construct nonconstant explicit solutions in terms of elementary (or even special) functions. – Start wearing purple Jun 24 '13 at 14:53
@O.L. Ah, yes, I see it now. – Douglas B. Staple Jun 24 '13 at 16:23

Maple yields this implicit solution: $$\int ^{y \left( x \right) }\!{\frac {1}{t\sqrt {2\,t-2\,y_{{0}}\ln \left( t \right) +B}}\;}{dt}=x+C$$ where $B,C$ are constants.

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Setting $u(t)=\ln\frac{y(x)}{y_0}$, $x=t/\sqrt{y_0}$, the equation for $u(t)$ becomes $$u''=e^u-1$$ Multiplying this by $2u'$ and integrating, we find $$(u')^2=2e^u-2u+A.$$ Subsequently separating the variables, one finds $$\int\frac{du}{\sqrt{2e^u-2u+A}}=\pm t+B,$$ where $A$, $B$ are two constants of integration.

P.S. This is equivalent to Maple answer of GEdgar.

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There is an easy implicit solution $$\frac{y}{y_0} = e^{ \frac{1}{6}\left( y-y_0 \right)^3 }.$$ Perhaps this can be converted into explicit solutions in terms of the Lambert-W function.

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This would give something for $y$ in terms of $y_0$ only, which would imply that $y(x)$ is a constant function of $x$. My suspicion is that the trivial solution $y=y_0$ is the only solution to follow from this equation. – Douglas B. Staple Jun 24 '13 at 14:03