Question

LAPLACE TRANSFORM METHOD

Answer 1

Use Laplace Transform method to solve the IVP y''+4y'+3y= (t when 0<t<2) and (0 when t>2) for y(0)=0 and y'(0)=0

Answer 2

define \(L(y(t))=Y(s)\) take laplace of both sides\[s^2Y-s y'(0)-y(0)+4(sY-y(0))+3Y=\int_{0}^{2} t e^{-st} \text{d}t=\frac{1-e^{-2s}(1+2s)}{s^2} \]\[(s^2+4s+3)Y=\frac{1-e^{-2s}(1+2s)}{s^2}\]\[Y(s)=\frac{1-e^{-2s}(1+2s)}{s^2(s^2+4s+3)}=\frac{1-e^{-2s}(1+2s)}{s^2(s+1)(s+3)}\]\[Y(s)=\frac{1}{s^2(s+1)(s+3)}-e^{-2s}\frac{1+2s}{s^2(s+1)(s+3)}\]

Answer 3

\[Y(s)=\color\red{\frac{1}{s^2(s+1)(s+3)}}-e^{-2s}\color\red{\frac{1+2s}{s^2(s+1)(s+3)}}\]partial fractioning for red parts and then using laplace inverse transform will lead u to the final answer, u have some heaviside step function because of that \(e^{-2s}\)