1. ## 2OLE (Simple)

$\displaystyle y'' - 2 y' + 1 y = -18 e^{1 t}$

Find a particular solution to this equation:

How exactly am I supposed to answer this?

My homologous equation renders repeated roots:

[Math] r_1=e^x, r_2=xe^x[/tex]

And when solve this with undetermined coefficients, I get:
$\displaystyle Y_p=Ae^t$

but since that e^x and xe^x already exist in the homologous solution, I should get:

$\displaystyle Y_p=\alpha T^2e^t$

Is this the correct answer? I cant solve for alpha, because I have no initial values missing. My assignment wont accept this format. What am I missing?

2. Originally Posted by Vamz
$\displaystyle y'' - 2 y' + 1 y = -18 e^{1 t}$

Find a particular solution to this equation:

How exactly am I supposed to answer this?

My homologous equation renders repeated roots:

[Math] r_1=e^x, r_2=xe^x[/tex]
Shouldn't this be
$\displaystyle y = Ae^t + Bte^t$

Originally Posted by Vamz
And when solve this with undetermined coefficients, I get:
$\displaystyle Y_p=Ae^t$

but since that e^x and xe^x already exist in the homologous solution, I should get:

$\displaystyle Y_p=\alpha T^2e^t$

Is this the correct answer? I cant solve for alpha, because I have no initial values missing. My assignment wont accept this format. What am I missing?
T is not the same variable as t. So your particular solution should be $\displaystyle y_p = \alpha t^2e^t$. Put your particular solution into the differential equation. I get $\displaystyle \alpha = -9$. So your overall solution will be
$\displaystyle y(t) = Ae^t + Bte^t - 9t^2 e^t$.

-Dan