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.


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...


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...


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 $Y$...



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...





Answer in Text:

No answer in text to compare this solution with.


I love solutions that are short, simple, and sweet as this but is it correct? Did you find any errors?


1 Answer 1


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 $t \lt 1$, $y(t) = 0$.

  • $\begingroup$ Gorden Thank you for your input but I find it confusing because you use advanced methods that I have not yet learned. For example although I know what integrating plus to minus infinity means and how to do that I do not know how this relates to this problem. Another concept we have not yet learned is complex integration (residue theorem). Please try again with a simpler method no more complex (difficult or rigorous) than what is given in the problem. $\endgroup$ Mar 6, 2015 at 23:48
  • $\begingroup$ This problem must be solved using convolution theorem. $\endgroup$ Mar 6, 2015 at 23:51
  • 1
    $\begingroup$ @JulesManson: you asked for a check on your answer. It has been provided. You erred in your penultimate step, very trivial. Otherwise what you did was good. My way provided an independent check. $\endgroup$
    – Ron Gordon
    Mar 6, 2015 at 23:54
  • $\begingroup$ I gave your answer credit because it was useful to a point. Again a bit too rigorous for my level of math. I really just wanted my method checked. Thank you again. $\endgroup$ Mar 7, 2015 at 0:08
  • $\begingroup$ To anyone interested: This post still needs help. I need someone to verify that my setup and the method (must be by convolution) for solving this problem has been done correctly. $\endgroup$ Mar 7, 2015 at 6:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .