1. ## Parametric differential equation

Hello everybody,
i'm really stuck with this and i hope somebody can help me out.

I have this differential equation:
$\displaystyle {y}'''-3{y}''+2{y}'+6y=e^{^{hx}}$
sadly there is no further explaination on what i should do with the parameter, so here is how i got through:
i wrote the omogeneous equation:
$\displaystyle {y}'''-3{y}''+2{y}'+6y=0$
then tried to find its general solution by writing the characteristic equation:
$\displaystyle \lambda^{3}-3\lambda^{2}+2\lambda+6=0$
$\displaystyle \left ( \lambda +1 \right )\left ( \lambda^{2}-4\lambda+6 \right )=0$
which has the following zeroes:
real:
$\displaystyle \lambda_{1}=-1$
and
complex:
$\displaystyle \lambda_{2,3}=2\pm i\sqrt{2}$
Which leads to the general solution of the omogeneous equation.
Now, on to the parameter; due to the fact that the $\displaystyle x$ coefficient, correct me if i'm wrong, changes the way my particular solution of the complete equation is wrote, should i study the cases where $\displaystyle h$ equals the characteristic equation zeroes? And if this is the right procedure, when i study the case in which:
$\displaystyle h=2+i\sqrt{2}$ or $\displaystyle h=2-i\sqrt{2}$
my solution will have the complex exponent?
Such as: $\displaystyle A*e^{2+i\sqrt{2x}}$ ?

Hope this is a clear enough question, excuse me for my poor english

No problem about a particular solution. According to a well known theorem, if $\displaystyle h$ is not a root of the characteristic equation a particular solution has the form $\displaystyle y(x)=Ce^{hx}$ . If $\displaystyle h$ is a root of the characteristic equation of multiplicity $\displaystyle s$, a particular solution has the form $\displaystyle y(x)=Cx^se^{hx}$ . Of course if $\displaystyle h\in\mathbb{R}$ you needn't consider the roots $\displaystyle 2\pm\sqrt{2}i$ .