Dirac Delta Function — Question 2

Question 2 – Solution

Strategy. Scale the delta first, then evaluate its smooth multiplier at the impulse time. The initial state contributes its own response area.

Step 1: Normalize and sift. Here $\delta(2t-4)=\delta(t-2)/2$ and $\delta(3-t)=\delta(t-3)$. The sampling property gives $$\boxed{f(t)=\tfrac 32\delta(t-2)+3\delta(t-3).}$$ The first weight is $(2^2-1)/2=3/2$. Reversing an argument does not turn a positive unit mass into a negative one: changing integration variables reverses the limits as well as introducing the negative derivative. Thus the scale factor is $1/|c|$.

Step 2: Transform and invert. The initial term is retained: $$(s+1)Y-2=\tfrac 32e^{-2s}+3e^{-3s},\qquad \boxed{Y=\frac{2+\tfrac 32e^{-2s}+3e^{-3s}}{s+1}.}$$ Consequently $$\boxed{y(t)=2e^{-t}+\tfrac 32H_2(t)e^{-(t-2)} +3H_3(t)e^{-(t-3)}.}$$

Step 3: Verify the jumps and intervals. Every active exponential satisfies $y'+y=0$ away from $2$ and $3$. The newly activated terms give $$[y]_2=\tfrac 32,\qquad [y]_3=3,\qquad y(0)=2.$$ Equivalently, integrating the equation across either impulse makes the bounded $y$ integral vanish as the interval shrinks, leaving the specified jump. Ordinary uniqueness on each interval and these jumps determine the whole response.

Step 4: Account for the initial stored response. Each shifted exponential of coefficient $C$ has integral $C$. Hence $$\boxed{\int_0^\infty y(t)\,dt=2+\tfrac 32+3=\tfrac{13}{2}.}$$ Integrating the impulsive equation over the half-line gives $y(\infty)-y(0)+\int_0^\infty y\,dt=9/2$ and the same result. The derivative contribution is $-2$, not zero. The positive nonzero $e^{-t}$ tail gives exact real domain $\boxed{s>-1}$.