Answer to Question #153825 in Differential Equations for Ra

Question

Obtain the inverse Laplace transform of t cos pt, p is not equal to 0.

Answer 1

$F(t)=tcos(pt),p\neq0\\ \therefore \mathcal{L}^{-1}\{F(s)\}\\ =\frac{d}{dt}\left(\mathcal{L}^{-1}\left\{\cos \left(ps\right)\right\}\right)+\mathcal{L}^{-1}\left\{\cos \left(ps\right)\right\}\left(0\right)$

Now $\mathcal{L}^{-1}\left\{\cos \left(ps\right)\right\} : No \; solution \;for\; standard \; function$

$\mathcal{L}^{-1}\{F(s)\}$ has no solution.

Now for Laplace transform,

$F(t)\\ =tcos(pt)\\ =\frac{1}{2p}2pcos(pt)\\ =\frac{1}{2p}\left[(sin(pt)+pcos(pt))-(sin(pt)-ptcos(pt))\right]$

$\\\;\\ \therefore \mathcal{L}\{F(t)\}\\ =\mathcal{L}\left(\frac{1}{2p}\left[(sin(pt)+pcos(pt))-(sin(pt)-ptcos(pt))\right]\right)\\ =\frac{1}{2p}\left[\mathcal{L}(sin(pt)+pcos(pt))-\mathcal{L}(sin(pt)-ptcos(pt))\right]\\ = \frac{1}{2p}\left[\frac{{2p{s^2}}}{{{{\left( {{s^2} + {p^2}} \right)}^2}}}-\frac{{2{p^3}}}{{{{\left( {{s^2} + {p^2}} \right)}^2}}}\right]\\ =\frac{1}{2p}\times \frac{{2ps^2-2{p^3}}}{{{{\left( {{s^2} + {p^2}} \right)}^2}}} =\frac{1}{2p}\times \frac{{2p(s^2-p^2)}}{{{{\left( {{s^2} + {p^2}} \right)}^2}}}\\ =\frac{s^2-p^2}{(s^2+p^2)^2}$

Therefore, Laplace transform of $\pmb{F(t)=\frac{s^2-p^2}{(s^2+p^2)^2}}$ .