Question

SOLUCIÓN DE ECUACIONES DIFERENCIALES CON TRANSFORMADA DE LAPLACE

dy/dt+ 2y=te^(-2t);y(0)=0

Answer 1

$y'+2y=te^{-2t}\\\\\mathcal{L}_s\{y'+2y\}=\mathcal{L}_s\{te^{-2t}\}\\\\(s+2)\mathcal{L}_s\{y\}-y(0)=\dfrac{1}{(s+2)^2}\\\\\\\mathcal{L}_s\{y\}=\dfrac{1}{(s+2)^3}\\\\\\y=\mathcal{L}^{-1}_t\left\{\dfrac{1}{(s+2)^3}\right\}\\\\\\\boxed{y=\dfrac{1}{2}t^2e^{-2t}}$