How do I solve Euler's equation? I have one:
$$x^2y''-4xy'+6y = 0$$
well, I know that substitution like $x = e^t$ works here and therefore $\ln x = t$
but I am not sure what to do next, as I can judge I should find derivatives to substitute instead of $y'$ and $y''$, but I am nit sure how to find them
 A: As @Dmoreno said you can use the ansatz $y = x^n$, this is the standard ansatz for the Euler-Cauchy Differential equation.
This will result in:
$$x^2n(n-1)x^{n-2}-4xnx^{n-1}+6x^n=0$$
$$\implies n(n-1)-4n+6=0$$
$$\implies n^2-5n+6=0 \implies (n-3)(n-2)=0$$
Solve the resulting quadratic equation to obtain two values for $n$. Let us call them $n_1=2$ and $n_2=3$ then the general solution is given by:
$$y(x)=c_1x^{n_1}+c_2x^{n_2}=c_1x^2+c_2x^3.$$
A: If $t= \ln(x)$ then $$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ and $\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)$$= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x}\frac{d}{dx}\frac{dy}{dt}$$= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$.
So $x^2\frac{d^2y}{dx^2}- 4\frac{dy}{dx}+ 6y= -\frac{dy}{dy}+ \frac{d^2y}{dt^2}- 4\frac{dy}{dt}+ 6y= \frac{d^2y}{dt^2}- 5\frac{dy}{dt}+ 6y= 0$, second differential equation with constant coefficients.  The characteristic equation is $r^2- 5r+ 6= (r- 3)(r- 2)= 0$ with roots $r= 2$ and $3$.  The two independent solutions to that equation are $e^{2t}$ and $e^{3t}$.
Since  $t= \ln(x)$  those become $e^{2\ln(x)}= e^{\ln(x^2)}= x^2$ and $e^{3ln(x)}= e^{\ln(x^3)}= x^3$.
The general solution to your original equation is $y(x)= C_1x^2+ C_2x^3$.
