## anonymous one year ago solve using laplace transforms y''+2y'+5y=(e^-t)sint

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1. IrishBoy123

got any IV's?

2. anonymous

use identity for laplace diffrential form

3. anonymous

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