Laplace doesn't apply in this case even though initial conditions have been provided, because the DE has variable coefficients: If Y" and Y' are replaced by a Laplace transform we get a new function which we can call y which is a function of s, namely y(s). But we still have the variable X in the equation instead of just y and s.
However, it may be possible to reduce the DE to a first degree DE. For this to work we need to be give (or to easily derive) one solution for the DE. No such solution has been given.
Let Y=Xn, then Y'=nXn-1, Y"=n(n-1)Xn-2, so the DE becomes:
n(n-1)Xn+nXn-3Xn=0, so n2-n+n-3=0, n2=3, n=±√3, so Y=X±√3.
So we can choose one solution Y=X√3 as a possible solution. At this stage it might be guessed that the final solution takes the form AX√3+BX-√3. Let's see if it is.
CHECK: Y=X√3, Y'=√3X√3-1, Y"=√3(√3-1)X√3-2, so X2Y"+XY'-3Y=(3-√3)X√3+√3X√3-3X√3=0 ✔️ Note that Y=AX√3 would also satisfy the DE, as would Y=X-√3.
So now it should be possible to reduce the order of the DE.
The solution above can be regarded as one solution for Y, and we can call it Y1. The next task is to find Y2 so that the final solution is Y=Y1+Y2.
To start this process let Y2=VY1=VX±√3. Knowing that both versions of Y1 are represented by Xn, let's make Y2=VXn for now.
Y2'=XnV'+nXn-1V, Y2"=XnV"+nXn-1V'+nXn-1V'+(n2-n)Xn-2V=XnV"+2nXn-1V'+(n2-n)Xn-2V.
Substitute this into the original DE:
x2(XnV"+2nXn-1V'+(n2-n)Xn-2V)+X(XnV'+nXn-1V)-3VXn=0,
Xn+2V"+2nXn+1V'+n2XnV-nXnV+Xn+1V'+nXnV-3XnV=0,
Xn+2V"+(2n+1)Xn+1V'+V(n2Xn-nXn+nXn-3Xn)=0. Remembering that n2=3, the highlighted expression is zero, so:
Xn+2V"+(2n+1)Xn+1V'=0. Let Z=V', then Z'=V". This reduces the order of the DE. Also Xn+1 is a common factor so it can be removed:
XZ'+(2n+1)Z=0, XZ'=-(2n+1)Z, which can be written:
dZ/Z=-(2n+1)dX/X. After integration we get: ln|Z|=-(2n+1)ln|X|+K where K is an integration constant. If K=ln(a):
ln|Z|=ln(aX-(2n+1)) and Z=aX-(2n+1).
But Z=dV/dX, so dV/dX=aX-(2n+1), V=b-aX-2n/(2n) and Y2=VY1=(b-aX-2n/(2n))Xn=bXn-aX-n/(2n), where b is another constant.
Y=Y1+Y2=Xn+bXn-aX-n/(2n)=BX√3-AX-√3 or BX-√3+AX√3 as expected. A and B absorb the constants a and b.
Since we already know that the exponent can be ±√3, Y satisfies the original DE, but the initial conditions cannot be applied, because we would have division by zero.