Table Of Laplace Transforms — Question 10

Question 10 – Solution

Strategy. Separate the initial response, delayed ordinary forcing and impulse response before inversion.

Step 1: Transform with all data. The left side transforms to $(s^2Y-s+1)+2(sY-1)+Y=(s+1)^2Y-s-1$. Consequently $$\boxed{Y=\frac 1{s+1}+\frac{e^{-s}}{(s+1)^4} +\frac{2e^{-3s}}{(s+1)^2}.}$$ The ordinary forcing has transform $e^{-s}/(s+1)^2$; the impulse has transform $2e^{-3s}$.

Step 2: Invert and make the intervals explicit. Define $v(\tau)=\tau^3e^{-\tau}/6$ and $h(\tau)=\tau e^{-\tau}$ for $\tau\ge 0$. The factorial in the fourth-power table row gives $$\boxed{y(t)=\begin{cases} e^{-t},&0\le t<1,\\ e^{-t}+v(t-1),&1\le t<3,\\ e^{-t}+v(t-1)+2h(t-3),&t\ge 3. \end{cases}}$$

Step 3: Verify each kind of activation. The initial branch gives $(y(0),y'(0))=(1,-1)$. For $D=d/d\tau$ and any polynomial $P$, $(D+1)^2[e^{-\tau}P(\tau)]=e^{-\tau}P''(\tau)$. Thus $v$ produces $\tau e^{-\tau}$ and $h$ produces zero between switches. At $1$, $v(0)=v'(0)=0$, so displacement and velocity remain continuous. At $3$, $h(0)=0$, $h'(0)=1$, so $[y]_3=0$ and $[y']_3=2$. Here $[z]_a=z(a^+)-z(a^-)$. These are exactly the required jump conditions.

Step 4: Check area, convergence and uniqueness. Direct areas are $1$, $3!/6=1$, and $2$, giving $\int_0^\infty y=4$. Independently, integrating the equation, including the velocity jump, gives $1-2+\int y=1+2$, again yielding $\boxed{\int y=4}$. The tail is a nonzero cubic polynomial times $e^{-t}$, so the exact real domain is $s>-1$. Linear IVP uniqueness on each ordinary interval, together with the forced matching and jump conditions, fixes the whole solution.