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One solution of $$x^2 y''' - 3x^2 y'' + 6xy' -6y = 0$$

for $x>0$ is $\phi_1 = x$. Find a basis for the solutions for $x>0$.

My attempt : I try to find $u(x)$ such that $u\phi_1$ is another solution. $$L(ux)=x^2(ux)'''-3x^2(ux)''+6x(ux)'-6(ux)$$

$$=x^3u'''+3x^2u''-3x^3u''$$

at this point let $u' = v$. We have the equation $$x^2v''+3x^2v'-3x^3v'=0$$ $$(x^3v')'-3x^3v' = 0$$

This is now a linear homogenous differential equation in $(x^3v')$, and I could go on to solve $v$ from here, which might give me one or two $u$ functions to be used for solutions of the original equations.

Can this method work and how should one approach higher order homogenous equations such as this? Doing all these product rules seems very excessive, and I'm wondering if there is a "smart way" I'm missing.

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  • $\begingroup$ Is the leading coefficient correct? If it were $x^3$ you would have an Euler-Cauchy DE. $\endgroup$ – Lutz Lehmann Apr 16 at 18:19
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$$x^2 y''' - 3x^2 y'' + 6xy' -6y = 0$$ Yo can easily reduce the order: $$ y''' - 3 y'' + 6\left(\dfrac yx\right)' = 0$$ Integrate. $$ y'' - 3 y' + 6\dfrac yx = C_1$$ $$ y'' - 3 (y' - 2\dfrac yx) = C_1$$

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