General Solution of $y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0,\; y_1= x$ (reduction of order method)

Question

I was given the question

Use the reduction of order method to find the general solution of the equation below. One solution of the homogeneous equation is shown alongside the equation $$ y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0,\quad y_1= x $$

Please these were the steps I took:

$y = y_1\int{u} \ dx $

where $u = u(x)$

$y = x\int{u} \ dx$

$y' = xu + \int{u} \ dx$

$y'' = xu' + 2u $

substituting $y,y',y''$ into $ y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0 $

$xu' + 2u - \frac{2}{x}(xu + \int{u}\ dx) + \frac{2}{x^2}x\int{u}\ dx $

$xu' + 2u -2u - \frac{2}{x}\int{u}\ dx + \frac{2}{x}\int{u}\ dx $

$xu' = 0 $

$x\ du = dx $

$ du = \frac1x\ dx $

$ u = \ln{x} + c $

substituting $$ u = \ln{x} + c $$ into $$y = y_1\int{u} \ dx $$

$ y = x \int{\ln{x} + c} $

$ y = x^2( \ln{x} - 1) + c_1x^2 + c_2x $

But the general solution was $$ y = c_1x + c_2x^2 $$

Please where did I go wrong or how do I get to the final solution of $ y = c_1x + c_2x^2 $

Answer 1

You made a mistake in solving $xu'=0$. This does not give you $xdu=dx$. The correct solution for this is $u=c$ a constant.