DIFFERENTIAL EQUATION?

x^2 y" + 3xy' + 4y = 0
y1(x) =x^2 (given)

Let y(x) = u(x) y1(x)
y = u (x^2)
y' = 2x u + (x^2)u'
y" = 2xu’ + 2u + (x^2)u” +2 x u’
y" = 4xu’ + 2u +(x^2)u”

Since
x^2 y" + 3xy' + 4y = 0
So
x^2 {4xu’ + 2u +(x^2)u”}+ 3x{ 2x u + (x^2)u'}+ 4u (x^2) = 0
therefore
x^4 (u") + x^3(u') = 0

let w = u' then w' = u"
x^4(w') + x^3(w )= 0
x^4 dw/dx = -x^3(w )
x^4 dw = -x^3(w) dx
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 (x)= {C ln x + C1} x^2

general solution : y(x) = C (x^2) (ln x ) + C1 (x^2)

2nd solution : y2(x) = (x^2) (ln x )

x^2*y" - 3*x*y' + 4*y = 0.
Set y = x^m.
Thus y' = m*x^(m-1) and y" = m(m-1)*x^(m-2).
Insert into the equation:
0 = [m*(m-1) - 3*m + 4]*x^m
= m^2 - 4m + 4
= (m - 2)^2
==> m = 2
==> y(x) = c*x^2 = Answer