Synopsis:
Please check my work. I do not have a text "answers to odd problems" for reference as this is an "even" numbered problem. The following documents in good detail the steps taken to solve for this so that the root of any errors, if any occur, can easily be found. Your input is very graciously welcomed.
Given:
Solve the following initial value problem where $r(t)=u(t-1)$ and $y(0)=0$ and $y'(0)=0$ by applying the convolution theorem…
$$y''+4y=r(1)$$
As a reference the convolution theorem formula is given and defined as…
$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$
My Solution:
We have a unit step function as the driving force $r(t)$ which implies…
$$r(t)=0 \implies [0<t<1)$$
$$r(t)=1 \implies [t\ge1)$$
Because $r(t)$ only exists when $t\ge1$ the convolution is now defined as…
$$h(t)=(f*g)(t)=\int_1^t f(\tau)g(t-\tau)d\tau$$
and the equation now looks like this…
$$y''+4y=1$$
We begin by finding the Laplace transform of $y''$…
$$\mathcal{L}\{y''\}=s^2\cdot Y-s\cdot y(0)-y'(0)=s^2Y$$
Now we are ready to replace $y''$ with $\mathcal{L}\{y''\}$ while taking the Laplace transform of entire equation and solving for $Y$…
$$Y(s^2+4)=\frac{1}{s}$$
$$Y=\frac{1}{s}\cdot\frac{1}{s^2+4}=\frac{1}{s}\cdot\frac{1}{2}\cdot\frac{2}{s^2+4}$$
This problem renders two transforms each with a related function as its inverse…
$$F(s)=\frac{1}{s} \implies f(t)=1$$
$$G(s)=\frac{1}{2}\cdot\frac{2}{s^2+4}\implies g(t)=\frac{1}{2}\sin(2t)$$
We now form the convolution by inserting the two associated functions (inverse transforms) into the formula $h(t)$ and simplify by switching $t$ and $\tau$ in the sine and changing its sign…
$$h(t)= 1*\frac{1}{2}\sin(2t)$$
$$=\frac{1}{2}\int_1^t 1\cdot\sin(2(t-\tau))d\tau$$
$$=-\frac{1}{2}\int_1^t \sin(2(\tau-t))d\tau$$
This yields…
$$y(t)=-\frac{1}{2}\left[-\frac{\cos(2\tau-t)}{2}\right]_1^t$$
$$=\frac{1}{4}\cos(2\tau-2t)\Big|_1^t$$
$$=\frac{1}{4}-\frac{1}{4}\cos(2(1-t))$$
$$=\frac{1}{4}-\frac{1}{4}\cos(2(t-1))$$
Answer in Text:
No answer in text to compare this solution with.
Question:
I love solutions that are short, simple, and sweet as this but is it correct? Did you find any errors?
Best Answer
The LT of the unit step is
$$\int_1^{\infty} dt \, e^{-s t} = \frac{e^{-s}}{s} $$
The LT of the equation is then
$$(s^2+4) Y(s) = \frac{e^{-s}}{s} \implies Y(s) = \frac{e^{-s}}{s (s^2+4)}$$
We find $y(t)$ by an inverse LT:
$$y(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, Y(s) e^{s t} = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{e^{(t-1) s}}{s (s^2+4)}$$
I apologize if this is not a familiar topic, but I will evaluate this using complex integration, i.e., the residue theorem. We have poles at $s=0$, $s=i 2$, and $s=-i 2$. By the residue theorem, $y(t)$ is the sum of the residues at the poles. Thus,
$$y(t) = \frac14 + \frac1{i 2 (i 4)} e^{i 2 (t-1)} + \frac1{-i 2 (-i 4)} e^{-i 2 (t-1)} = \frac12 \sin^2{(t-1)}$$
This is valid for $t \gt 1$. When $t \lt 1$, $y(t) = 0$.