y''+9y=30sin(3x)
I'm told to use the method of solving y''+9y=30e^(3ix), but I'm not sure what to do after I find (D-3)(D-3)y=30e^(3ix).
Could I have some help?
In terms of Laplace Transform the particular integral of...
$\displaystyle y^{''} + 9\cdot y = 30 \cdot \sin 3x$ (1)
... is written as...
$\displaystyle Y_{p} (s) = \frac{10}{(s^{2}+9)^{2}} $ (2)
... so that is...
$\displaystyle y_{p} (x) = \mathcal {L} \{Y_{p} (s)\} = \frac {5}{27}\cdot (\sin 3x - 3x\cdot \cos 3x)$ (3)
Alternatively [and is is a not confortable job ...] You can search for function of the type...
$\displaystyle y_{p} (x) = x \cdot (\alpha\cdot \cos 3x + \beta\cdot \sin 3x)$ (4)
... that satisfies (1)...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$