<s>[INST] Second order nonhomgeneous equation euler stuck at nonhomogeneous part there is second order nonhomogeneous differential equation, euler 
Im stuck at find nonhomogeneous equation,
$x^2y"-3xy'+4y=log x$  and i have to find the solution
my attempt : 
first i found solution for the homogeneous equation, it is a repeated root so
$yp=Ax^2lnx+Bx^2$
here for the nonhomgenous i used variation parameter
$y1= x^2ln x$ and $y2=x^2$
using Wronski and cramer rule: 
$yp=u1.y1+u2.y2$  
W=$ \left[
  \begin{array}{ c c }
    x^2lnx & x^2 \\
     2xlnx+x & 2x
  \end{array} \right] $
$=-x^3$ 
W1=$ \left[
  \begin{array}{ c c }
     0 & x^2 \\
     log x & 2x
  \end{array} \right]$
$=-x^2lnx$

W2=$ \left[
  \begin{array}{ c c }
     x^2lnx & 0 \\
     2xlnx+x & log x
  \end{array} \right]$
$=x^2ln^2x$ 
$u'=\frac{W_1}{W}\frac{-x^2lnx}{-x^3}=\frac{lnx}{x}$
$u'=\frac{W_2}{W}\frac{x^2ln^2x}{-x^3}=\frac{-ln^2x}{x}$ 
integrate to find  u1, u2 
$u1=\frac{1}{2}ln^2x$ 
$u1=-\frac{1}{3}ln^3x$ 
$yp=u1.y1+u2.y2$  
$=\frac{1}{2}ln^2x$$.x^2lnx$ $-\frac{1}{3}ln^3x$ $.x^2$
$=x^2ln^3x\frac{1}{6}$ 
 is this right??? 
however the book said that the answer is $y=x^2(Alogx+B)+\frac{1}{4}(logx+1)$
why this is not the same?? 
 [/INST] I believe your error was in finding $u_1^{\prime}$ and $u_2^{\prime}$;
you should have
$\displaystyle\frac{\ln x}{x^2}$ instead of logx in the expressions
for $W1$ and $W2$, so that $u_1^{\prime}=-\frac{(\ln x)^2}{x^3}$ and $u_2^{\prime}=\frac{\ln x}{x^3}$.
This will give $u_1=\left(\frac{1}{2}(\ln x)^2+\frac{1}{2}\ln x+\frac{1}{4}\right)(x^{-2})$ and $u_2=-\left(\frac{1}{2}\ln x+\frac{1}{4}\right)(x^{-2})$, so
$y_{p}=u_1y_1+u_2y_2=\frac{1}{2}(\ln x)^2+\frac{1}{2}\ln x+\frac{1}{4}-\frac{1}{2}(\ln x)^2-\frac{1}{4}\ln x=\frac{1}{4}\left(\ln x+1\right)$.

If you use the method of undetermined coefficients instead, with $y_p=A\ln x+B$,
you will find that $A=\frac{1}{4}$ and $B=\frac{1}{4}$.
</s>