Using Laplace transforms calculate the solution of the ordinary differential equation (d^2 y)(dt^2)-2 (dy)(dt)+y=e^(-t), subject to the initial conditions y(0)=0 and (dy)(dt)(0)=1. Select the correct answer from the following list. a) y(t)=(3te^t)/(2)+(e^(-t))/(4)-(e^t)/(4) b) y(t)=te^t +(t^2 e^t)/(2) c) y(t)=(te^(-t))/(2)-(e^(-t))/(4)+(e^t)/(4) d) y(t)=te^(-t)+(t^2 e^(-t))/(2) e) y(t)=(te^(-t))/(2)-(e^(-t))/(2)+(e^t)/(2)

Question

Using Laplace transforms calculate the solution of the ordinary differential equation

\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} + y = e^{- t}

, subject to the initial conditions y(0)=0 and

\frac{d y}{d t} (0) = 1

. Select the correct answer from the following list.

a) y (t) = \frac{3 t e^{t}}{2} + \frac{e^{- t}}{4} - \frac{e^{t}}{4} b) y (t) = t e^{t} + \frac{t^{2} e^{t}}{2} c) y (t) = \frac{t e^{- t}}{2} - \frac{e^{- t}}{4} + \frac{e^{t}}{4} d) y (t) = t e^{- t} + \frac{t^{2} e^{- t}}{2} e) y (t) = \frac{t e^{- t}}{2} - \frac{e^{- t}}{2} + \frac{e^{t}}{2}

Answer 1

\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} + y = e^{- t} y (0) = 0, y^{'} (0) = 1

Auxillary equation and given by

m^{2} = 2 m + 1 = 0 \Rightarrow (m - 1)^{2} = 0 \Rightarrow m = 1, 1

complementary solution is

y_{2} (x) = (c_{1} + c_{2} t) e^{t}

where

c_{1}

and

c_{2}

are some constant
Particular integral

\frac{1}{D^{2} - 2 D + 1} e^{- t} = \frac{1}{(- 1)^{2} - 2 (- 1) + 1} e^{- t} = \frac{1}{4} e^{- t}

complete solution (y)

y (t) = (c_{1} + c_{2} t) e^{t} + \frac{1}{4} e^{- t} + \frac{1}{4} e^{- t}

using integral conclusion

y (0) = c_{1} + \frac{1}{4} \Rightarrow c_{1} = - \frac{1}{4} y^{'} (t) = (c_{1} + c_{2} t) e^{t} + e^{t} \times c_{2} - \frac{1}{4} e^{- t} y^{'} (0) = c_{1} e^{0} + c_{2} - \frac{1}{4} e^{- 0} \Rightarrow c_{1} + c_{2} - \frac{1}{4} \Rightarrow y (t) = - \frac{1}{4} e^{t} + \frac{3}{2} t e^{t} + \frac{1}{4} e^{- t}

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