a. Integrate both sides of the equation to obtain

\int_{x(0-)}^{x(t)} dx = \int_{0-}^{t} \delta (t) dt = \int_{0-}^{0+} \delta (t) dt + \int_{0+}^{t} \delta (t) dt = 1+0

because the area under a unit impulse is 1. This gives

x(t)=x(0-)=1 \quad \text{or} \quad x(t)=x(0-)+1=3+1=4

This is the solution for t > 0 but not for t = 0−. Thus, x(0+) = 4 but x(0−) = 3, so the impulse has changed x(t) instantaneously from 3 to 4.

b. The transformed equation is

sX(s)-x(0-)=1 \quad \text{or} \quad X(s)=\frac{1+x(0-)}{s}

which gives the solution x(t) = 1 + x(0−) = 4. Note that the initial value used with the derivative property is the value of x at t = 0−

The initial-value theorem gives

x(0+)=\underset{s \rightarrow \infty}{\text{lim}} s X(s)= \underset{s \rightarrow \infty}{\text{lim}} s \frac{1+x(0-)}{s}=1+3=4

which is correct.