Can you solve a linearly independent ODE's homogeneous part without the particular solution. For example $y''-6y'+9y=xe^{3x},\hspace{5mm} y(0)=0,y'(0)=0 $
I was wondering if it is possible to compute the homogenous solution even before working out the particular solution to this equation.
As an aside, I was wondering if the initial conditions are (0,0) for y' and y then is it not the case that I don't need the particular solution. As, in effect I am only lookin at the homogenous side of the equation.
I think my misunderstanding comes form the fact that in $y_{c}=y_{p}+y_{h}$ i see $y_{p}$ as a translation or rotation of the solution space away from the origin, and therefore if the inital condition are centered at (0,0) then I know the solution must exist in $y_{h}$.

Solving the problem $$y''-6y'+9y=xe^{3x},\hspace{5mm} y(0)=0,y'(0)=0$$
as if it was a homogenous problem and plugging in indeed results in the actual (IVP-solved) general solution to the inhomogenous problem.
General solution:
$$y(x)=\frac{1}{6}e^{3x}x^3+e^{3x}c_1+e^{3x}c_2$$
IVP-General solution:
$$y(x)_{IVP}=\frac{1}{6}e^{3x}x^3$$
A counter-example would be:
$$y''-6y'+9y=e^{3x}+cos(x),\hspace{5mm} y(0)=0,y'(0)=0$$
Where the general solution is:
$$y(x)=\frac{2}{25}\cos(x)-\frac{3}{50}\sin(x)+e^{3x}\left(\frac{1}{2}x^2+\frac{1}{2}c_1x+\frac{1}{2}c_2\right)$$
The IVP-general solution on the other hand is:
$$y(x)_{IVP}=\frac{2}{25}\cos(x)-\frac{3}{50}\sin(x)+e^{3x}\left(\frac{1}{2}x^2+\frac{15}{50}x-\frac{4}{50}\right)$$
In which the constants $c_1$ and $c_2$ are clearly not zero.
