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Given that $y=x$ is a solution, use the method of reduction of
order to find the general solution of :
$x^{3} y^{\prime \prime \prime}-3 x^{2} y^{\prime \prime}+x\left(6-x^{2}\right) y^{\prime}-\left(6-x^{2}\right) y=0$

$y=x \rightarrow y=v x$

$y^{\prime}=v^{\prime} x+v$

$y^{\prime \prime}=v^{\prime \prime} x+v^{\prime}+v^{\prime}=v^{\prime \prime} x+2 v^{\prime}$

$y^{\prime \prime \prime}=v^{\prime \prime \prime} x+v^{\prime \prime}+2 v^{\prime \prime}=v^{\prime \prime \prime} x+3 v^{\prime \prime}$

$\rightarrow$ Substitute in the D.E

$x^{3}\left(v^{\prime \prime \prime} x+3 v^{\prime \prime}\right)-3 x^{2}\left(v^{\prime \prime} x+2 v^{\prime}\right)+x\left(6-x^{2}\right)\left(v^{\prime} x+v\right)-\left(6-x^{2}\right)(v x)=0$

$x^{4} v^{\prime \prime \prime}+3 x^{3} v^{\prime \prime}-3 x^{3} v^{\prime \prime}-6 x^{2} v^{\prime}+\left(6 x-x^{3}\right)\left(v^{\prime} x+v\right)-6 v x+x^{3} v=0$

$x^{4} v^{\prime \prime \prime}+3 x^3 v^{\prime \prime}-3 x^{3} v^{\prime \prime}-6 x^{2} v^\prime+6 x^{2} v^\prime+6 x v-x^{4}v^\prime-x^3v-6 v x+x^{3} v=0$

$x^{4} v^{\prime \prime \prime}-x^{4} v^\prime=0 \quad \div x^{4}$

$v^{\prime \prime \prime}-v^{\prime}=0$

$\left(D^{3}-D\right) v=0 \longrightarrow f(m)=0$

$m^{3}-m=0 \rightarrow m\left(m^{2}-1\right)=0 \quad m-$ values: $\quad 0,+1,-1$

$v=c_{1} e^{0}+c_{2} e^{-x}+c_{3} e^{+ x} \quad$ But $\quad y=v x$

$y=\left(c_{1} e^{0}+c_{2} e^{-x}+c_{3} e^{x}\right) x$

Where $c_{1}, c_{2}, c_{3}$ are arbitrary constants.