I have a linear homogeneous equation whose characteristic equation has the following roots:
$$
\alpha, \beta, \gamma, \gamma, \gamma, \delta +i \epsilon, \delta + i \epsilon, \delta – i \epsilon, \delta – i \epsilon
$$
That's nine roots, both real and complex, some repeated, and the complex ones in conjugate pairs.
What is the general solution of this equation?
If the roots are
$$
\alpha, \beta, 0, 0, 0, \delta +i \epsilon, \delta + i \epsilon, \delta – i \epsilon, \delta – i \epsilon
$$
how does this modify the solution?
This was a serious lacuna in the maths I did for my physics degree and I'm trying to get myself back up to speed now I have a daughter who is studying this stuff.
Best Answer
Assuming the independent variable is $t$, the general solution can be written as $$ c_1e^{\alpha t}+c_2e^{\beta t}+e^{\gamma t}(c_3+c_4t+c_5t^2) +e^{\delta t}(c_6\cos(\epsilon t)+c_7\sin(\epsilon t)) +te^{\delta t}(c_8\cos(\epsilon t)+c_9\sin(\epsilon t)). $$ If $\gamma=0$, it becomes $$ c_1e^{\alpha t}+c_2e^{\beta t}+c_3+c_4t+c_5t^2 +e^{\delta t}(c_6\cos(\epsilon t)+c_7\sin(\epsilon t)) +te^{\delta t}(c_8\cos(\epsilon t)+c_9\sin(\epsilon t)). $$