Solving IVP’s with Laplace Transforms — Question 7

Question 7 – Solution

Strategy. Shift the independent variable and the forcing together. A Laplace transform beginning at the new origin must use data at that origin.

Step 1: Move the initial time. With $\tau=t-1$, the derivatives have the same form but the forcing becomes $\tau+1$. Thus $$z''+z=\tau+1,\qquad z(0)=2,\qquad z'(0)=0.$$ Writing $Z=\mathcal L_\tau\{z\}$ gives $$(s^2+1)Z-2s=\frac 1{s^2}+\frac 1s.$$

Step 2: Resolve and invert. A useful decomposition is $$Z=\frac{2s+1/s^2+1/s}{s^2+1} =\frac 1{s^2}+\frac 1s+\frac{s}{s^2+1}-\frac 1{s^2+1}.$$ Therefore $\boxed{z(\tau)=\tau+1+\cos\tau-\sin\tau}$. Transforming each term back checks the decomposition, including the constant part of the shifted forcing.

Step 3: Return to the original clock. The solution on the requested domain is $$\boxed{y(t)=t+\cos(t-1)-\sin(t-1),\qquad t\ge 1.}$$ Its first derivative is $1-\sin(t-1)-\cos(t-1)$ and its second is $-\cos(t-1)+\sin(t-1)$. Hence $y''+y=t$, while $y(1)=2$ and $y'(1)=0$. These are the actual specified data, so uniqueness on $[1,\infty)$ verifies the result.

Step 4: Diagnose the two clock errors. The values two and zero belong to $z(0),z'(0)$, not to values of $y$ at an unspecified earlier time. Once that clock change is made, the forcing is $\tau+1$ and its transform is $1/s^2+1/s$, not just $1/s^2$. The transform being inverted is $Z$, defined using $\tau\ge 0$. Since $z(\tau)=\tau+O(1)$ is eventually positive, it has exact domain $\boxed{s>0}$. No extension of $y$ to times before one was needed.