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$x^4y'' + c^2y = 0$: How would you go on about solving this problem? WolframAlpha solved it by substituting $y=x \cdot v(x)$ and then letting $t=\frac{1}{x}.$

Does anyone have a more intuitive way of solving it? If not, why is the substitution exactly $y=x \cdot v(x)$? However, I have never heard of this method before. What I tried to do is naively let $x^2=u$, and then proceed to do the equation as an Euler-Cauchy equation. I came to a solution but it is nothing alike the one WolframAlpha provided which leads me to think it can't be solved that way.

amWhy
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l0ner9
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    $y=xv(x)$ is a cheap way to knock the $x^4y''$ down to a more reasonable $x^2v''+\text{lower derivatives}$. If you substitute $x^2=u$ you get an equation of the form $(\alpha uD_u^2+\beta D_u+\gamma)y=0$ which isn't Euler-Cauchy but regular singular point $u=0$ for Frobenius method. – user10354138 Sep 06 '21 at 12:47
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    Mostly, the equations are not solvable by hand. If it is solvable one way is enough and we should feel thankful. – Z Ahmed Sep 06 '21 at 15:08
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    Unfortunately, almost no differential equations have "intuitive" solutions. Usually finding any way to solve them, intuitively or not, is enough of a gift. – K.defaoite Sep 06 '21 at 15:19
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    The solution is in the book:"DIFFERENTIALGLEICHUNGEN LOSUNGSMETHODEN UND LOSUNGEN" E. KAMKE. – giuseppe mancò Sep 06 '21 at 18:32
  • Related: https://math.stackexchange.com/questions/1804280/verify-y-xaz-p-leftbxc-right-is-a-solution-to-y-left-frac1-2ax-ri. – Gonçalo Jul 15 '24 at 02:20
  • Recent related question about $x^4y''-y=0$: https://math.stackexchange.com/questions/4942460/how-to-solve-linear-differential-equation-with-polynomial-coefficients – Lutz Lehmann Jul 16 '24 at 04:46

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