Finding the general solution

Question

$y''+\frac{2}{x}y' = \frac{1}{x^2}$

Homo. Solution: $y= A+B\frac{1}{x}$

Part. Solution: Using Lagrange's method:

$y_p=A(x)+B(x)\frac{1}{x}$

Assumption:

$A'(x)+B'(x)\frac{1}{x} = 0 \rightarrow A'(x) = -B'(x) \frac{1}{x} (1)$

$A'(x)*0 - B'(x)\frac{1}{x^2} = \frac{1}{x^2} \rightarrow B'(x) = -1\ (2)$

$\Rightarrow B = -x$

Add (2) to (1) $\Rightarrow A'(x) = \frac{1}{x} \Rightarrow A = ln(x)$

Part. Solution: $y_p=ln(x) - 1$

I believe I made a mistake somewhere. May you show me?

Answer 1

HINT

I would say simplify

$\displaystyle y''+\frac{2}{x}y' = \frac{1}{x^2}$

$\displaystyle y'=p, y''=p'\Rightarrow p'+\frac{2}{x}p = \frac{1}{x^2}\Rightarrow p = c_1\frac{1}{x^2}+\frac{1}{x}=y'\Rightarrow y =\ln x -\frac{C_1}{x}+C_2$