How can $x^2y''-3xy'+3y=0$ be solved? a) How does $x^2y''-3xy'+3y=0$ can be solved? I know how to solve for constant coefficients, but in this case they are functions...
b) In which maximum interval there is a solution that  confirms $y'(1)=6, y(1)=4$?
 A: Hint: This is Euler-Cauchy type equation, so let:
$$y = x^m \implies y' = m x^{m-1} \implies y'' = m(m-1)x^{m-2}$$
Substitute back into ODE, solve for roots, write the general solution and then find constants from initial conditions.
Spoiler

 $\qquad \qquad \qquad \qquad y(x) = x(x^2 + 3)$

A: An equation of the form $ax^2u''+bxu'+cu=0$ can be rewritten in terms of the operator $D=x\frac{d}{dx}$: indeed, we have $$ax^2u''+bxu'+cu=aD^2u+(b-a)Du+cu.$$
The right hand side is the result of applying the operator $aD^2+(b-a)D+c$ to $u$. If we are able to factor $aD^2+(b-a)D+c=\alpha(D-\beta)(D-\gamma)$, then we are left with trying to solve the equation $$(D-\beta)(D-\gamma)u=0.$$. Now it is clear that if we manage to find a function such that $$(D-\gamma)u=0$$ we will have a solution, and this equation is $xu'=\gamma u$, which can easily be solved because its variables are separable. Likewise, we can find a solution to out original problem by looking for solutions of $$(D-\beta)u=0.$$ If $\beta\neq\gamma$ in this way we obtain two linearly independent solutions to our original equation.
A: $$am^2+(b-a)m+c=0\\
m^2-4m+3=0\\
(m-3)(m-1)=0\\
m_1=3, \, m_2=1\\
m_1 \neq m_2 \text{ so}\\
Y=C_1X^3+C_2X^1   \text{ is general solution}$$
