Question

Resultado de la ecuacion de laplace y''-y'=e^t*cos*t, y(0)=0, y'(0)=0 , .

Answer 1

$y''-y'=e^t \cos t\\ \\ \mathcal{L}\{y''-y'\}(s)=\mathcal{L}\{e^t \cos t\}(s)\\ \\ \mathcal{L}\{y''\}(s)-\mathcal{L}\{y'\}(s)=\mathcal{L}\{\cos t\}(s-1)\\ \\ \left[s^2\mathcal{L}\{y\}(s)-sy(0)-y'(0)\right]-\left[s\mathcal{L}\{y\}(s)-y(0)\right]=\dfrac{s-1}{(s-1)^2+1}\\ \\ \\ s^2\mathcal{L}\{y\}(s)-s\mathcal{L}\{y\}(s)=\dfrac{s-1}{(s-1)^2+1}\\ \\ \\ s(s-1)\mathcal{L}\{y\}(s)=\dfrac{s-1}{(s-1)^2+1}$


$\mathcal{L}\{y\}(s)=\dfrac{1}{s[(s-1)^2+1]}\\ \\ \\ \mathcal{L}\{y\}(s)=\dfrac{1/2}{s}+\dfrac{1}{(s-1)^2+1}-\dfrac{s/2}{(s-1)^2+1}\\ \\ \\ y=\mathcal{L}^{-1}\left\{\dfrac{1/2}{s}+\dfrac{1}{(s-1)^2+1}-\dfrac{s/2}{(s-1)^2+1}\right\}(t)$


$y=\dfrac{1}{2}\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\}+\mathcal{L}^{-1}\left\{\dfrac{1}{(s-1)^2+1}\right\}-\dfrac{1}{2}\mathcal{L}^{-1}\left\{\dfrac{s}{(s-1)^2+1}\right\}\\ \\ \\ y=\dfrac{1}{2}+\dfrac{1}{2}\mathcal{L}^{-1}\left\{\dfrac{1}{(s-1)^2+1}\right\}-\dfrac{1}{2}\mathcal{L}^{-1}\left\{\dfrac{s-1}{(s-1)^2+1}\right\}\\ \\ \\ y=\dfrac{1}{2}+\dfrac{1}{2}e^t\sin t-\dfrac{1}{2}e^t\cos t\\ \\ \\ \boxed{y=\dfrac{1}{2}+\dfrac{1}{2}e^t(\sin t-\cos t)}$