Solving IVP’s with Laplace Transforms — Question 2

Question 2 – Solution

Strategy. Linearity separates the numerator contribution of initial data from that of the input, but the final answer needs both.

Step 1: Transform all derivative terms. The derivative transforms are $s^2Y-s+1$ and $sY-1$. Thus $$(s^2+3s+2)Y-s-2=4/s,$$ $$Y=\underbrace{\frac{s+2}{(s+1)(s+2)}}_{Y_{\mathrm{initial}}} +\underbrace{\frac 4{s(s+1)(s+2)}}_{Y_{\mathrm{forced}}}.$$ The first part simplifies to $1/(s+1)$; the second resolves as $2/s-4/(s+1)+2/(s+2)$.

Step 2: Invert and combine. Hence $$y_{\mathrm{initial}}=e^{-t},\qquad y_{\mathrm{forced}}=2-4e^{-t}+2e^{-2t},$$ $$\boxed{y(t)=2-3e^{-t}+2e^{-2t}.}$$ The initial-data response satisfies the homogeneous equation, and the constant term in the forced response supplies the constant forcing.

Step 3: Verify each contribution. For the sum, $y'=3e^{-t}-4e^{-2t}$ and $y''=-3e^{-t}+8e^{-2t}$. Then $$y''+3y'+2y=4+(-3+9-6)e^{-t}+(8-12+4)e^{-2t}=4.$$ Also $y(0)=1$ and $y'(0)=-1$. The initial-data part has values $1,-1$ and zero equation residual. The forced part has values $0,0$ and residual four. These checks establish the intended decomposition; uniqueness of the linear IVP establishes the final solution.

Step 4: Interpret the steady value and domain. The exponentials vanish, so $\boxed{y(t)\to 2}$. This agrees with the constant balance $2y=4$. Its ordinary transform converges exactly for $s>0$: boundedness gives convergence above zero, while the eventually positive nonzero constant tail causes divergence at or below zero. Although the initial-data response has the wider domain $s>-1$, the complete response does not. Reporting only the forced response gets the limiting value right but misses both prescribed initial data.