a. Let v(t) = ẋ(t). Then equation (1) becomes v̇= δ(t), which can be integrated to obtain v(t) = v(0−) + 1 = 10 + 1 = 11. Thus, ẋ(0+) = 11 and is not equal to ẋ(0−). Now integrate ẋ = v = 11 to obtain x(t) = x(0−) + 11t = 5 + 11t. Thus, x(0+) = 5, which is the same as x(0−).

So for this model the unit-impulse input changes ẋ from t = 0− to t = 0+ but does not change x.

b. The transformed equation is

s^2X(s)-sx(0-)-\overset{.}{x}(0-)=1

or

X(s)=\frac{sx(0-)+\overset{.}{x}(0-)+1}{s^2}=\frac{5s+11}{s^2}=\frac{5}{s}+\frac{11}{s^2}

which gives the solution x(t) = 5 + 11t and ẋ(t) = 11. Note that the initial values used with the derivative property are the values at t = 0−

The initial-value theorem gives

x(0+)=\underset{s \rightarrow \infty}{\text{lim}} sX(s)= \underset{s \rightarrow \infty}{\text{lim}} s \frac{5s+11}{s^2}=5

and because \mathcal{L}(\overset{.}{x})=sX(s)-x(0-),

\overset{.}{x}(0+)= \underset{s \rightarrow \infty}{\text{lim}} s[sX(s)=x(0-)]= \underset{s \rightarrow \infty}{\text{lim}} s (\frac{5s+11}{s}-5)=11

as we found in part (a).