Question

Define a solution the given initial-value problem. 4'' + 4y'4y=delta(t - 4) y(0) = 1 y'(0) = 2

Transformation properties
ANSWERED
asked 2020-12-17
Define a solution the given initial-value problem.
\(4''\ +\ 4y'4y=\delta(t\ -\ 4)\)
\(y(0) = 1\)
\(y'(0) = 2\)

Answers (1)

2020-12-18

To solve, we use the properties of the Laplace transform and the Laplace table
The Laplace transform satisfies the linearity properties
\(L[f = g] = L[f]\ +\ L[g]\)
\(L[cf] = cL[f]\)
For all transfotmable functions f and g and constants c.
\(L[y''\ +\ 4y'\ +\ 4y] = L[\delta(t\ -\ 4)]\)
\(L[y'']\ +\ 4L[y']\ +\ 4L{y} = L[\delta(t\ -\ 4)]\)
\(s^{2}Y(s)\ -\ sy(0)\ -\ y'(0)\ +\ 4(sY(s)\ -\ y(0))\ +\ 4Y(s)=e^{-4s}\)
\(Y(s)(s^{2}\ +\ 4s\ +\ 4)\ -\ y'(0)\ -\ y(0)(s\ +\ 4)=e^{-4s}\)
Substitute initial conditions,
\(y(0) = 1\)
\(y'(0) = 2\)
Solve for \(Y (s).\)
\(Y(s) (s^{2}\ +\ 4s\ +\ 4)\ -\ 2\ -\ s\ -\ 4 = e^{-4s}\)
\(Y(s)=\frac{e^{-4s}\ +\ s\ +\ 6}{(s\ +\ 2)^{2}}\)
\(=\frac{e^{-4s}}{(s\ +\ 2)^{2}}\ +\ \frac{1}{s\ +\ 2}\ +\ \frac{4}{(s\ +\ 2)^{2}}\)
Inverse transform:
\(L^{-1}[Y(s)]=L^{-1}\left[\frac{e^{-4s}}{(s\ +\ 2)^{2}}\ +\ \frac{1}{s\ +\ 2}\ +\ \frac{4}{(s\ +\ 2)^{2}}\right]\)
\(y(t)=L^{-1}\left[e^{-4s}\ \frac{1}{(s\ +\ 2)^{2}}\right]\ +\ L^{-1}\left[\frac{1}{s\ +\ 2}\right]\ +\ 4L^{-1}\left[\frac{1}{(s\ +\ 2)^{2}}\right]\)
\(=L^{-1}[e^{-4s}L[te^{-2t}]]\ +\ e^{-2t}\ +\ 4te^{-2t}\)
\(=u_{4}(t)(t\ -\ 4)e^{-2(t\ -\ 4)}\ +\ e^{-2t}(1\ +\ 4t)\)

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