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Homedifferential equations for $x^x$

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# differential equations for $x^{x}$

In this entry, we will derive differential equations satisfied by the function $x^{x}$.
^{1}^{1}In this entry, we restrict $x$, and hence $x^{x}$ to be strictly positive
real numbers, hence it is justified to divide by these quantities.
We begin by computing its derivative. To do this, we write $x^{x}=e^{{x\log x}}$ and
apply the chain rule:

${d\over dx}x^{x}={d\over dx}e^{{x\log x}}=e^{{x\log x}}(1+\log x)=x^{x}(1+\log x)$ |

Set $y=x^{x}$. Then we have $y^{{\prime}}/y=1+\log x$. Taking another derivative, we have

${d\over dx}\left({y^{{\prime}}\over y}\right)={1\over x}\cdot$ |

Applying the quotient rule and simplifying, this becomes

$yy^{{\prime\prime}}-(y^{{\prime}})^{2}-y^{2}/x=0.$ |

It is also possible to derive an equation in which $x$ does not appear. We start by noting that, if $z=1/x$, then $z^{{\prime}}+z^{2}=0$. If, as above, $y=x^{x}$, we have $(d/dx)(y^{{\prime}}/y)=z$. Combining equations,

${d^{2}\over dx^{2}}\left({y^{{\prime}}\over y}\right)+\left({d\over dx}\left({% y^{{\prime}}\over y}\right)\right)^{2}=0;$ |

applying the quotient rule and simplifying,

$y^{3}y^{{\prime\prime\prime}}-y^{2}(y^{{\prime\prime}})^{2}+2y(y^{{\prime}})^{% 2}y^{{\prime\prime}}-3y^{2}y^{{\prime}}y^{{\prime\prime}}-(y^{{\prime}})^{4}+2% y(y^{{\prime}})^{3}=0.$ |

## Mathematics Subject Classification

26A99*no label found*

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