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My problem is this given Cauchy-Euler equation:

$$x^{3}y^{\prime \prime \prime} +xy^{\prime}-y=0$$

My approach: this is an differential equation, so i was looking for a solution with the method of undetermined coefficients. but honestly, i failed.

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Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^m\to y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0\Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $x\neq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+\ln x+\ln^2 x), ~x>0$$

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    $\begingroup$ $\quad \langle +\rangle_+^+\quad \bf \ddot\smile\;$ $\endgroup$
    – amWhy
    Jun 16 '13 at 1:07
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Hint: Use substitute $t=\log x$ for $x>0$ and $t=\log (-x)$ for $x<0$

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