Help with solving differential equations I tried to solve this third order differential equation:
$$x^3y’’’+2x^2y’’-xy’+y=0$$
By substituting $y=e^{kx}$ and then finding $k$.
But I am not sure if I should solve this by using this method?
 A: Hint:
$$x^3y'''+2x^2y''-xy'+y=(x^3y'''+3x^2y'')-(x^2y''+2xy')+(xy'+y)=(x^3y''-x^2y'+xy)'=0$$
then
$$x^2y''-xy'+y=\dfrac{C}{x}$$
let $x=e^t$ then the DE reduces to
$$Y''-2Y'+Y=Ce^{-t}$$
A: To solve the third order Cauchy Euler equation we start by making the substitution
$y=x^r$ 
then
$y'=rx^{r-1}, y''=r(r-1)x^{r-2}$ and $y'''=r(r-1)(r-2)x^{r-3}$
Substituting this back into the original differential equation yields,
$x^3r(r-1)(r-2)x^{r-3}+2x^2r(r-1)x^{r-2}-xrx^{r-1}+x^r=0$,
$x^r(r(r-1)(r-2)+2r(r-1)-r+1)=0$,
$r^3-3r^2+2r+2r^2-2r-r+1=0$,
$r^3-r^2-r+1=0$ or $r^2(r-1)-(r-1)=0$ or $(r^2-1)(r-1)=0$
so the roots are $r_1=-1$ of multiplicity $1$ 
and $r_2=1$ of multiplicity $2$.  Hence the general equation is
$y=c_1x^{-1}+c_2x + c_3x \ln{x}$
A: $$x^3y’’’+2x^2y’’-xy’+y=0\qquad . . . . . . . (1)$$
Let $~x=e^z\implies z=\log x$
Then $\quad y'=\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{x}\frac{dy}{dz}\implies xy'= \frac{dy}{dz}\equiv Dy\qquad \text{where} \quad D\equiv \frac{d}{dz}$
$y''=\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x}\frac{d^2y}{dz^2}\implies x^2y''=D(D-1)y$
Similarly $\quad x^3~y'''=D(D-1)(D-2)y$
Now $(1)$ becomes, $$\{D(D-1)(D-2)+2D(D-1)-D+1\}y=0$$
$$\implies (D-1)\{D(D-2)+2D-1\}y=0$$
$$\implies (D-1)^2(D+1)y=0$$
Roots of the auxiliary equation are $~1, ~1, ~-1$.
Hence the general solution is $$y=(c_1+~c_2~ z)e^z+c_3e^{-z}$$i.e.,
$$y=(c_1+c_2 \log x)~x+c_3~x^{-1}$$where $~c_1, ~c_2, ~c_3~$ are arbitrary constants.
