Solving IVP’s with Laplace Transforms — Question 9

Question 9 – Solution

Strategy. Higher derivatives introduce more initial terms. A repeated transform pole packages the polynomial-exponential solution compactly.

Step 1: Transform the derivative hierarchy. The general third-derivative rule is $\mathcal L\{y^{(3)}\}=s^3Y-s^2y(0)-sy'(0)-y''(0)$. Here the first three derivatives transform to $sY-1$, $s^2Y-s$ and $s^3Y-s^2$. Hence $$(s^3Y-s^2)+3(s^2Y-s)+3(sY-1)+Y=0,$$ $$\boxed{Y=\frac{s^2+3s+3}{(s+1)^3}.}$$ Even the zero initial values must be placed in the correct derivative formulas before simplifying.

Step 2: Expand at the repeated pole. Writing $p=s+1$ makes the numerator $p^2+p+1$, so $$Y=\frac 1{s+1}+\frac 1{(s+1)^2}+\frac 1{(s+1)^3},\qquad \boxed{y=e^{-t}(1+t+t^2/2).}$$ We have $y'=-t^2e^{-t}/2$ and $y''=e^{-t}(t^2/2-t)$. Thus the three initial values are exactly $1,0,0$.

Step 3: Verify the operator identity. For $y=e^{-t}v$, repeated product rules give $$y^{(3)}+3y''+3y'+y=e^{-t}v^{(3)}.$$ Here $v=1+t+t^2/2$ has third derivative zero. This verifies the equation everywhere. The linear third-order IVP with all three data has a unique solution.

Step 4: Check shape, area and domain. The polynomial factor is positive on $t\ge 0$, and $y'<0$ for every $t>0$, so $y$ is positive and strictly decreasing after its initially flat point. It tends to zero. Direct exponential moments give $$\boxed{\int_0^\infty y(t)\,dt=1+1+1=3=Y(0).}$$ The exact transform domain is $s>-1$; at or below $-1$ the positive polynomial tail makes the integral diverge. The data $y'(0)=y''(0)=0$ describe only the initial point. The equation gives $y^{(3)}(0)=-1$, so the first change is cubic rather than absent.