This is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version:

$$y''-3y'+2y=0$$

Solving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$.

So, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions.

To use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is:

$c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.

We now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$.

$$c_1'=e^t \implies c_1=e^t$$

$$c_2'=-te^t \implies c_2=-e^t(t-1)$$

Plugging these expressions back to $Y_{gen.hom}$ yields the solution which is

$y=te^t$