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

and thanks in advance,

SkyWolf.