Question

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

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

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:
$$L^{-1}[Y(s)]=L^{-1}\left[\frac{e^{-4s}}{(s\ +\ 2)^{2}}\ +\ \frac{1}{s\ +\ 2}\ +\ \frac{4}{(s\ +\ 2)^{2}}\right]$$
$$y(t)=L^{-1}\left[e^{-4s}\ \frac{1}{(s\ +\ 2)^{2}}\right]\ +\ L^{-1}\left[\frac{1}{s\ +\ 2}\right]\ +\ 4L^{-1}\left[\frac{1}{(s\ +\ 2)^{2}}\right]$$
$$=L^{-1}[e^{-4s}L[te^{-2t}]]\ +\ e^{-2t}\ +\ 4te^{-2t}$$
$$=u_{4}(t)(t\ -\ 4)e^{-2(t\ -\ 4)}\ +\ e^{-2t}(1\ +\ 4t)$$