I've been given the full solution
$L_{inhom} = c_1*e^{\lambda_1*t}+c_2*e^{\lambda_2*t}+ x_{01}(t)+x_{02}(t)$
Where $f(x_{0i})=q_i(t)$
I wish to find the coefficients for the original equation.
I know, quite circumstantially, that for higher order linear differential equation with real constants (HLDiff from now on) and degree 2, that
$(\lambda -\lambda _1)*(\lambda -\lambda _2)= \lambda ^2+a_1*\lambda +a_0$
I've found this by setting
$\lambda_1 = (-a_1-sqrt(a_1^2+4*a_0)/2 \wedge \lambda_1 = (-a_1+sqrt(a_1^2+4*a_0)/2$
And then expanding the expression.
I have tried it with other numerical lambdas and full solutions, where $\forall\lambda(\lambda\in\Re\wedge am(\lambda)=1)$, and it seems that the formula
$(\lambda -\lambda _1)*(\lambda -\lambda_2)+...+(\lambda-\lambda_{n})=\lambda^n+a_{n-1}*\lambda^{n-1}+...+a_0$
Works in a general sense.
Problem is, how would I go about proving this?
The solution I use (and therefore the knowledge I have wrt. HLDiffs) is dependent on turning the HLDiff of order n into a 1. Degree systems of linear differentialequations, and from there using the "diagonalization" formula:
$X(t) = c_1*e^{\lambda_1*t}*V_{\lambda_1} + ... + c_n*e^{\lambda_n*t}*V_{\lambda_n}$
With the lambdas found from the matrix that occurs as a result of turning the single equation into a system of equations.