<s>[INST] 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? 
 [/INST] 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$
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