Solving IVP’s with Laplace Transforms — Question 10

Question 10 – Solution

Strategy. Transform the target first, then use the differential operator to recover the necessary input. Check the permitted input family before selecting parameters.

Step 1: Transform the prescribed trajectory. The exponential and time-exponential pairs give $$Y_k=\frac 1s-\frac 1{s+k}-\frac{k}{(s+k)^2} =\boxed{\frac{k^2}{s(s+k)^2}}.$$ All targets have $y_k(0)=y_k'(0)=0$, so their required input transform is $$Q_k=(s^2+3s+2)Y_k=\frac{k^2(s+1)(s+2)}{s(s+k)^2}.$$

Step 2: Recover the input and impose the restriction. Partial fractions, verified by clearing denominators, give $$Q_k=\frac 2s+\frac{k^2-2}{s+k}-\frac{k(k-1)(k-2)}{(s+k)^2},$$ $$q_k(t)=2+(k^2-2)e^{-kt}-k(k-1)(k-2)te^{-kt}.$$ For $k>0$, the last term vanishes exactly for $\boxed{k=1\text{ or }k=2}$. Their respective values of $C$ are $-1$ and $2$. A nonzero $te^{-kt}$ coefficient cannot be absorbed into a constant multiple of $e^{-kt}$ on an interval.

Step 3: Select and verify the desired acceleration. Differentiation gives $y_k'=k^2te^{-kt}$ and $y_k''=k^2(1-kt)e^{-kt}$, so $y_k''(0)=k^2$. The added requirement selects $$\boxed{k=2,\quad C=2,\quad q=2+2e^{-2t},\quad y=1-(1+2t)e^{-2t}.}$$ Here $y'=4te^{-2t}$, $y''=(4-8t)e^{-2t}$, and $y''+3y'+2y=2+2e^{-2t}$. The initial values are $0,0$ and the acceleration is four, as required. Both $Y$ and $Q$ have exact domain $s>0$ because their original functions approach positive constants.

Step 4: Prove the rise and uniqueness of the input. Since $y'(t)>0$ for $t>0$, the response increases from zero. Also $1-y=(1+2t)e^{-2t}>0$ for every finite $t$, and this difference tends to zero. Thus $\boxed{0<y(t)<1\ (t>0),\quad y(t)\to 1}$ with no overshoot. Once the entire twice-differentiable target is fixed, the equation forces $q=y''+3y'+2y$ pointwise. The input is unique, rather than merely one successful choice; linear-IVP uniqueness then verifies that it produces this trajectory.