Here is the problem:
y''-10y'+25y = -5e^(5t) / (t^2 + 1)
Find the particular solution.
I am looking at the problem and does not know how to approach it at all
Please help!!
Thank you!!
As in a previous case we can try to write the differential equation in term of LT...
$\displaystyle s^{2}\cdot Y(s) -10\cdot s\cdot Y(s) + 25\cdot Y(s) = G(s)$ (1)
... where...
$\displaystyle G(s)= \mathcal{L} \{ \frac{-5\cdot e^{5t}}{1+t^{2}}\}$ (2)
... so that the 'particular solution' You are searching is...
$\displaystyle y^{*}(t) = \mathcal{L}^{-1} \{\frac{G(s)}{(s-5)^{2}} \}$ (3)
Till this point it is 'all right', but the problems start now. Setting...
$\displaystyle \Gamma(s)= \mathcal {L} \{\frac{1}{1+t^{2}}\}$ (4)
... it is easy to find that is...
$\displaystyle G(s)= -5\cdot \Gamma (s-5)$ (5)
... all right!... but if You loock at the 'manual' You find that is...
$\displaystyle \Gamma (s) = \cos s \cdot \{\frac{\pi}{2} - Si (s)\} - \sin s\cdot Ci(s)$ (6)
... where $\displaystyle Si (*)$ and $\displaystyle Ci(*)$ are the socalled 'Integralsine' and 'Integralcosine' functions... and the problem seems to become 'a little unconfortable' ...
May be some errors of me... or some errors of some other...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$