Find the base of the kernel of $L(y)=x^2y''-3xy'+3y.$ 
Let $L:C^2(I)\rightarrow C(I), L(y)=x^2y''-3xy'+3y.$ Find the kernel of the linear transformation $L$. Can the solution of $L(y)=6$ be expressed in the form $y_H$+$y_L$, where $y_H$ is an arbitrary linear combination of the elements of ker L.

What I have tried:
Since ker L is subspace of $C^2(I)$ and dim $C^2(I)=2$, dim ker $L\leq 2$.
Let $y(x)=x^r$. Then substituting gives $L(x^r)=x^rr^2-3rx^r+3x^r$, hence  $L(x^r)=0$ iff $x^rr^2-3rx^r+3x^r=0.$ $r$ can be solved using the quadratic formula.
$$r=\frac{3+i\sqrt3}{2} \vee  r=\frac{3-i\sqrt3}{2}$$ 
$$y_1(x)=x^\frac{3+i\sqrt3}{2}, y_1(x)=x^\frac{3-i\sqrt3}{2} $$which are linearly independent(?).
How would one show that $y_1$ and $y_2$ are LI? 
 A: The second derivative of $x^r$ is $r(r-1)x^{r-2}$. So the characteristic equation you should get is $$ r^2-4r+3=0,$$ with roots $1$ and $3$. This gives you a fundamental set of solutions
$$
y_1(x)=x,\ \ y_2(x)=x^3. 
$$
Because your equation is linear homogeneous of order two, any solution will be a linear combination of your fundamental solutions (this is important as your argument that $C^2(I)$ has dimension two is wrong; it's dimension is infinite). In other words, the kernel of $L$ is 
$$
\{\alpha x+\beta x^3:\ \alpha,\beta\in\mathbb R\},
$$
and it has dimension $2$. 
To know that $y_1$ and $y_2$ are a fundamental set of solutions, you need to verify that they are linearly independent. In general, you would do it with the Wronskian. But, for two functions, linear independence simply means that one is not a multiple of each other. As $x^3/x$ is not constant, they are linearly independent. 
A: Another way
$$x^2y''-3xy'+3y=0$$
For $x \neq 0$
$$y''-3 \left (\frac {xy'-y}{x^2} \right )=0$$
$$y''-3\left (\frac {y}{x}\right )'=0$$
$$y'-3 \left (\frac {y}{x} \right )=K_1$$
$$x^3y'-3{y}{x^2}=K_1x^3$$
$$\frac {x^3y'-3{y}{x^2}}{x^6}=\frac {K_1}{x^3}$$
$$\left (\frac {y}{x^3} \right )'=\frac {K_1}{x^3}$$
$$\frac {y}{x^3}=K_1\int\frac {dx}{x^3}=\frac {K_1}{x^2}+K_2$$
$$\boxed {y=K_1x+K_2x^3}$$
Both functions $(x,x^3)$ are linearly independant.
Show that $ \forall x \,\, c_1x+c_2x^3=0 \implies c_1=c_2=0$
