Question

solve using laplace transforms
y''+2y'+5y=(e^-t)sint

Answer 1

got any IV's?

Answer 2

use identity for laplace diffrential form

Answer 3

\(y" +2y' +5y= e^{-t}sin t\)

\(\mathcal{L}(y"+2y'+5y) = \mathcal {L}(e^{-t}sin(t)) \)

The left hand side :
\(\mathcal{L}(y") = s^2\mathcal {L}(y) -sy(0)-y'(0) \\\mathcal{L}(y')=s\mathcal{L}(y)-y(0)\)

The \(RHS = \dfrac{1}{(s+1)^2 +1}\). The expression becomes


\(s^2\mathcal {L}(y) -sy(0)-y'(0) +2\mathcal{L}(y)-2y(0) + 5\mathcal{L}(y)= \dfrac{1}{(s+1)^2 +1}\)
From here, you need y(0) and y'(0) to solve for \(\mathcal {L}(y)\)