Differential equation $y'' - \frac{2y}{x^2} = 3\ln(-x)$ Hello there I am trying to solve
$$y'' - \frac{2y}{x^2} = \ln(-x)$$
I get this differential equation from book - filippov differential equations link on book (problem #606).
And same way I have solve (in this book)
$$y = \frac{C_1}{x} + \left[ C_2 - \frac{1}{3}\ln(-x) + \frac{1}{2}\ln^2(-x) \right] x^2$$
But I can not get this solve.
My solve:
$$y'' - \frac{2y}{x^2} = \ln(-x)$$
$$x^2y'' - 2y = x^2 \ln(-x)$$
$$k(k-1) - 2 = 0$$
$$k_1 = -1, k_2= 2, y_1=x^{-1}, y_2=x^2$$
$$y_0 = C_1y_1 + C_2y_2 = \frac{C_1}{x} + C_2x^2$$
$$W = \begin{vmatrix}
y_1 &y_2 \\
y'_1 &y'_2
\end{vmatrix} = \begin{vmatrix}
x^{-1} &x^2 \\
-x^{-2} &2x
\end{vmatrix} = 2+1 = 3$$
$$W_1 = \begin{vmatrix}
0  &  y_2 \\
g(x) & *
\end{vmatrix} = \begin{vmatrix}
0 & x^2 \\
\ln(-x) & 2x
\end{vmatrix} = -x^2 \ln(-x)$$
$$W_2 = \begin{vmatrix}
y_1   & 0 \\
*     &g(x)  
\end{vmatrix} = \begin{vmatrix}
x^{-1} &0 \\
-x^{-2} & \ln(-x)
\end{vmatrix} = \frac{\ln(-x)}{x}$$
$$y_h = u_1y_1 + u_2y_2$$
$$u_1 = \int{\frac{W_1}{W}} = \int{\frac{-x^2 \ln(-x)}{3}} = -\frac{1}{9}x^3(3\ln(-x) - 1)$$
$$u_2 = \int{\frac{W_2}{W}} = \int{\frac{3\ln(-x)}{x}} = \frac{3}{2}\ln^2(-x)$$
$$y_h = -\frac{1}{9}x^3(3\ln(-x) - 1) x^{-1} + \frac{3}{2}\ln^2(-x) x^2 = -\frac{1}{3}x^2 \ln(-x) - \frac{x^2}{9} + \frac{3}{2}\ln^2(-x) x^2$$
$$y = y_0 + y_h$$
$$y = \frac{C_1}{x} + C_2x^2 - \frac{1}{3}x^2 \ln(-x) - \frac{x^2}{9} + \frac{3}{2}\ln^2(-x) x^2$$
General solution:
$$y = \frac{C_1}{x} + \left[ C_2 - \frac{1}{3} \ln(-x) - \frac{1}{9} + \frac{3}{2}\ln^2(-x) \right] x^2$$
Please help me where I'm wrong?
May be in book have wrong?
 A: HINT
You can also approach it as follows:
\begin{align*}
y'' - \frac{2y}{x^{2}} = \ln(-x) & \Longleftrightarrow x^{2}y'' - 2y = x^{2}\ln(-x)\\\\
& \Longleftrightarrow (x^{2}y'' + 2xy') - (2xy' + 2y) = x^{2}\ln(-x)\\\\
& \Longleftrightarrow (x^{2}y')' - (2xy)' = x^{2}\ln(-x)
\end{align*}
Can you take it from here?
A: I found error (thank you very much everyone in comment)
First error in differential equation:
$$y'' - \frac{2y}{x^2} = 3 \ln(-x)$$
Solution:
$$x^2y'' - 2y = x^2 3\ln(-x)$$
$$k(k-1) - 2 = 0$$
$$k_1 = -1, k_2= 2, y_1=x^{-1}, y_2=x^2$$
$$y_0 = C_1y_1 + C_2y_2 = \frac{C_1}{x} + C_2x^2$$
$$g(x) = 3 \ln(-x)$$
$$W = \begin{vmatrix}
y_1 &y_2 \\
y'_1 &y'_2
\end{vmatrix} = \begin{vmatrix}
x^{-1} &x^2 \\
-x^{-2} &2x
\end{vmatrix} = 2+1 = 3$$
$$W_1 = \begin{vmatrix}
0  &  y_2 \\
g(x) & *
\end{vmatrix} = \begin{vmatrix}
0 & x^2 \\
3\ln(-x) & 2x
\end{vmatrix} = -x^2 3 \ln(-x)$$
$$W_2 = \begin{vmatrix}
y_1   & 0 \\
*     &g(x)  
\end{vmatrix} = \begin{vmatrix}
x^{-1} &0 \\
-x^{-2} & 3\ln(-x)
\end{vmatrix} = \frac{3\ln(-x)}{x}$$
$$y_h = u_1y_1 + u_2y_2$$
$$u_1 = \int{\frac{W_1}{W}} = \int{\frac{-x^2 3\ln(-x)}{3}} = \int{-x^2 \ln(-x)} = - \frac{1}{9}x^3(3\ln(-x) - 1) = - \frac{1}{3}x^3\ln(-x) + \frac{x^3}{9}$$
$$u_2 = \int{\frac{W_2}{W}} = \int{\frac{\frac{3\ln(-x)}{x}}{3}} = \int{\frac{\ln(-x)}{x}} = \frac{1}{2}\ln^2(-x)$$
$$y_h = (- \frac{1}{3}x^3\ln(-x) + \frac{x^3}{9}) x^{-1} + \frac{1}{2}\ln^2(-x) x^2 = -\frac{1}{3}x^2 \ln(-x) + \frac{x^2}{9} + \frac{1}{2}\ln^2(-x) x^2$$
$$y = y_0 + y_h$$
$$y = \frac{C_1}{x} + C_2x^2 -\frac{1}{3}x^2 \ln(-x) + \frac{x^2}{9} + \frac{1}{2}\ln^2(-x) x^2$$
$$y = \frac{C_1}{x} + [C_2 - \frac{1}{3} \ln(-x) + \frac{1}{9} + \frac{1}{2}\ln^2(-x)] x^2$$
And because $\frac{1}{9}$ - constant, we can combine with $C_2$
General solution:
$$y = \frac{C_1}{x} + [C_2 - \frac{1}{3} \ln(-x) + \frac{1}{2}\ln^2(-x)] x^2$$
