Parametric differential equation

**SkyWolf** · Sep 2nd 2011, 11:29 AM

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

I have this differential equation:
${y}'''-3{y}''+2{y}'+6y=e^{^{hx}}$
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:
${y}'''-3{y}''+2{y}'+6y=0$
then tried to find its general solution by writing the characteristic equation:
$\lambda^{3}-3\lambda^{2}+2\lambda+6=0$
$\left ( \lambda +1 \right )\left ( \lambda^{2}-4\lambda+6 \right )=0$
which has the following zeroes:
real:

$\lambda_{1}=-1$

and
complex:

$\lambda_{2,3}=2\pm i\sqrt{2}$

Which leads to the general solution of the omogeneous equation.
Now, on to the parameter; due to the fact that the $x$ 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 $h$ equals the characteristic equation zeroes? And if this is the right procedure, when i study the case in which:
$h=2+i\sqrt{2}$ or $h=2-i\sqrt{2}$
my solution will have the complex exponent?
Such as: $A*e^{2+i\sqrt{2x}}$ ?

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

and thanks in advance,
SkyWolf.

**FernandoRevilla** · Sep 2nd 2011, 12:03 PM

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

**SkyWolf** · Sep 2nd 2011, 12:11 PM

Thanks a lot man.

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