Particular solution of non-homogeneous second-order ODE $x^{2}y''-3xy'+4y=x^{2}\ln(x)$ Given equation
$$x^{2}y''-3xy'+4y=x^{2}\ln(x)$$
Calculated homogeneous equation would be 
$$y_h=(a+bz)e^{2z}$$
and the particular solution,
$$y_p=\frac{1}{(D-2)^{2}}ze^{2z}$$
How do I have to process the double integration for the particular solution?
 A: An easy way for solving this Euler-Cauchy' differential equation. Rewrite it as:
$$(e^{-2t}y)''=t$$
Where $t=\ln x$
.Integrate twice:
$$y(t)=e^{2t} \left (c_1+c_2t+\frac 16 t^3 \right )$$
Unsubstitute $t= \ln x$ :
$$\boxed {y(x)=x^2 \left (c_1+c_2\ln x+\frac 16 \ln^3 x\right )}$$

But if you prefer to use the D-Operator then:
$$y_p=\frac{1}{(D-2)^{2}}ze^{2z}=e^{2z}\frac{1}{D^{2}}z$$
$$y_p=e^{2z}\frac{1}{D}\dfrac {z^2}2=e^{2z}\dfrac {z^3}6$$
$$ \implies y_p=x^2\dfrac {\ln^3 x}6$$
A: $$Y''-\frac{3}{x}Y'+\frac{4}{x^2}Y= \ln x=f(x)~~~(1)$$
Let us first solve the homogeneous part:
$$y''-\frac{3}{x}y'+\frac{4}{x^2}y=0$$
It is Euler's equation we use: $y=x^m$, to get
$$m(m-1)-3m+4=0 \implies m=2,2$$
So $y_1=C_1 x^2, C_2 x^2 \ln x ~~~(2)$. For the solution of (1), we have to vary $C_1, C_2$
Then $$Y(x)= C_1(x) x^2+ C_2(x)x^2 \ln x~~~(2)$$
Where $$C_1(x)=-\int \frac{f(x) y_2(x)}{w} dx+D_1, ~~C_2(x)=\int \frac{f(x) y_1(x)}{w} dx+D_2.$$
Here $w(x)=x^3$ is the wronskian of $y_1(x), y_2(x)$
$$C_1(x)=-\int \frac{\ln^2 x}{x} dx +D_1= -\frac{\ln^3 x}{3}+D_1~~~(3)$$
Similarly, $$C_2(x)=\int \frac{\ln x}{x}+D_2=\frac{\ln^2 x}{2}+D_2=~~(4)$$ 
Inserting (3) and (4) in (2) we get the total solution of (1) as
$$Y(x)=\frac{x^2 \ln^3 x}{6}+D_1x^2+D_2x^2\ln x.$$
