given $y_1 = x$ and the equation: $x^3y'''-3x^2y''+(6-x^2)xy'-(6-x^2)y = 0$

I determined $y_2 = ux$,

$y_2' = u+xu'$,

$y_2''= 2u'+xu''$,

and $y_2'''= 3u''+xu'''$

substitution in the equation gives me: $x^4u'''-x^4u' = 0$

then, let $v=u'$, $v'=u''$, and $v''=u'''$ gives me $x^4v'' - x^4v = 0$

how do you go about solving after that? i believe it has to do with separable equations, but i can not figure it out. HELP PLEASE!!!

  • $\begingroup$ You have $v''=v$, which has solution $C_1e^{-x} + C_2e^x$. $\endgroup$ – Chee Han Dec 1 '16 at 6:45

You properly did the work and assuming $x\neq 0$, your last equation write $$v''-v=0$$ which cannot be reduced anymore.

The characteristic equation being $r^2-1=0\implies r=\pm 1$, then the general solution is given by $$v=c_1 e^x+c_2 e^{-x}$$ which makes $$u=c_1 e^x-c_2 e^{-x}+c_3$$ and $$y=x(c_1 e^x-c_2 e^{-x}+c_3)$$ SInce we use constants, change $-c_2$ to $c_2$.


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