Find the general solution and 2nd solution of x^2 y'' + 3x y' + y = 0 y1(x) = 1 / x (given)?

x^2 y" + 3xy' + y = 0
y1(x) =1/x (given)

Let y(x) = u(x) y1(x)
y = u/x
y' = u (-x^-2) + (x^-1)u'
y" = -u' (-x^-2) + (2x^-3) u + (x^-1)u" - u'(x^-2)
y" = (2x^-3)u - 2(x^-2)u + (x^-1) u"
Since
x^2 y" + 3xy' + y = 0
So
x^2 { (2x^-3)u - 2(x^-2)u + (x^-1) u"} + 3x{u (-x^-2) + (x^-1)u' }+ u/x = 0

therefore
-2u' + xu" + 3u' = 0
xu" + u' = 0
let w = u'
then w' = u"
xw' + w = 0
xw' = -w
x dw/dx = -w
(1/w) dw = (-1/x) dx
int (1/w) dw = int (-1/x) dx
ln w = -1 ln x + c
e^(ln w) = e^( -1 ln x + c)
w = e^( ln x^ -1) ( e^ c) = (x^-1 ) C
w = u'
du/dx = (x^-1 ) C
du = (x^-1 ) C dx
int du = int (x^-1 ) C dx
u (x) = C ln x + C1
y = u x^-1 = {C ln x + C1} x^-1

general solution :
y(x) = C (ln x ) / x + C1/x

2nd solution : (ln x ) / x

y' = 1 / x
y"=-1/x^2
x^2 y'' + 3x y' + y = 0
(x^2)(-1/x^2)+3x/x +y=0
-1+3+y=0
y=-2