c28 - 18.03 Class 28 Apr 21 Laplace Transform III Second...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
18.03 Class 28, Apr 21 Laplace Transform III: Second order equations; completing the square. Rules: L is linear: L[af(t) + bg(t)] = aF(s) + bG(s) F(s) essentially determines f(t) s-shift: L[e^{at}f(t)] = F(s-a) t-shift: L[f_a(t)] = e^{-as} F(s) s-derivative: L[tf(t)] = - F'(s) t-derivative: L[f'(t)] = s F(s) - f(0+) L[f"(t)] = s^2 F(s) - s f(0+) - f'(0+) (ignoring singularities at t = 0 ) Computations: L[1] = 1/s L[e^{as}] = 1/(s-a) L[cos(omega t)] = s/(s^2+omega^2) L[sin(omega t)] L[delta(t-a)] = = omega/(s^2+omega^2) e^{-as} L[t^n] = n!/s^{n+1} , n = 1, 2, 3, ... [1] To handle second degree equations we'll need to know the LT of f"(t). We'll compute it by regarding f"(t) as the derivative of f'(t). We'll employ a technique here that will get repeated several more times today: pick a new function symbol and use it to name some function that arises in the middle of a calculation. The application of this principle here is to write g(t) = f'(t) . We'll assume that f(t) and f'(t) are continuous for t > 0 and ignore singularities at t = 0 , so that G(s) = L[g(t)] = s F(s) - f(0+) Write down the t-derivative rule for g(t): L[g'(t)] = s G(s) - g(0+) So then
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
L[f"(t)] = s (s F(s) - f(0+)) - f'(0+) = s^2 F(s) - s f(0+) - f'(0+) [2] Example: x" + 2 x' + 5 x = 5 , x(0+) = 2, x'(0+) = 3 . Note that this is EASY to solve using our old linear methods: by inspection (or undetermined coefficients, or the Key Formula) xp = 1/5 is a solution; the general solution is this plus a homogeneous solution, which you choose to satisfy the initial conditions.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern