Given the following equation:
$x^2y''\:-\:3xy'\:+\:3y\:=\:lnx$
When I let $lnx$ be $z$, I am not able to understand the transition that yields:
$y''\:-\:4y'\:+3y\:=\:z$
I tried using things I know:
$z = lnx$
$e^z = x$
$x^2 = e^{2z}$
but I am not able to get the transition yet.
The equations defined by operators of the form $L(y)=\sum_{k=0}^n a_kx^ky^{(k)}=b(x)$, with $a_k\in\mathbb R$ can be solved applying the substitution $y(x)=y(e^t)$. Observe that $L(y)=b(x)$ is not in normal form so you have to study the equation in $J=(-\infty,0)\cup(0,+\infty)$.
Consider $x\in(0,+\infty)$, if $y(x)=y(e^t)=Y(t)$, you obtain $Y'(t)=y'(e^t)e^t=\boxed{y'(x)x}$ and $Y''(t)=y''(e^t)e^te^t+y'(e^t)e^t=y''(e^t)e^{2t}+Y'(t)$. From these relations if follows that $Y''(t)-Y'(t)=y''(e^t)e^{2t}=\boxed{y''(x)x^2}$.
The initial equation is now $Y''(t)-Y'(t)-3Y'(t)+3Y(t)=\log(e^t)$, so $$Y''(t)-4Y'(t)+3Y(t)=t$$ which is a second order differential equation with constant coefficients and the general solution is given by $Y(t)=Y_{hom}(t)+Y_p(t)$, where $Y_{hom}$ is the solution for the homogeneous problem and $Y_p(t)$ is the particular solution.
The characteristic polynomial associated to the homogeneous problem is $P(\mu)=\mu^2-4\mu+3\mu$, whose roots are $\mu_{1,2}=\dfrac{4\pm2}{2}=2\pm1$, then $$Y_{hom}(t)=c_1e^{3t}+c_2e^{t}, \hspace{1cm}c_1,c_2\in\mathbb R.$$ The general particular solution is $Y_p(t)=at+b$, with $a,b\in\mathbb R$ to determine.