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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.

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