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 $y'(1)=6, y(1)=4$?
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


   $\qquad \qquad \qquad \qquad y(x) = x(x^2 + 3)$
$$am^2+(b-a)m+c=0\\
m^2-4m+3=0\\
(m-3)(m-1)=0\\
m_1=3, \, m_2=1\\
m_1 \neq m_2 \text{ so}\\
Y=C_1X^3+C_2X^1   \text{ is general solution}$$