Solving IVP’s with Laplace Transforms — Question 3

Question 3 – Solution

Strategy. A repeated operator root and a matching input pole combine. Pole order controls the polynomial factor, not necessarily growth of the full time response.

Step 1: Form the transformed equation. The initial terms give $$(s^2Y-2s+2)+2(sY-2)+Y=\frac 1{s+1},$$ so $$(s+1)^2Y=2(s+1)+\frac 1{s+1},\qquad Y=\frac 2{s+1}+\frac 1{(s+1)^3}.$$ The forcing contributes a third-order pole at $s=-1$, although the differential equation has order two.

Step 2: Invert the repeated pole correctly. Since $\mathcal L\{t^2e^{-t}\}=2/(s+1)^3$, $$\boxed{y(t)=e^{-t}(2+t^2/2).}$$ The factor $1/2$ is essential. Transforming the two terms back gives exactly the displayed $Y$.

Step 3: Check the residual and data. With $v=2+t^2/2$, the product rule gives $y'=e^{-t}(v'-v)$ and $y''=e^{-t}(v''-2v'+v)$. Therefore $$y''+2y'+y=e^{-t}v''=e^{-t}.$$ At zero, $v(0)=2$, $v'(0)=0$, so $y(0)=2$ and $y'(0)=-2$. These checks and linear-IVP uniqueness verify the solution.

Step 4: Distinguish a scaled response from the actual response. We have $$y'=e^{-t}(-2+t-t^2/2) =-\tfrac 12e^{-t}[(t-1)^2+3]<0.$$ Thus $y$ decreases strictly and tends to zero, since exponential decay dominates its quadratic factor. Meanwhile $\boxed{e^t y=2+t^2/2\to\infty}$. Dividing out the decaying envelope changes the quantity being studied; there is no contradiction. The positive polynomial-exponential tail gives the exact domain $\boxed{s>-1}$, with divergence at and below the boundary.