1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Y" -y' -2y = 4(t^2) , y(0) = 1 , y'(0) =4?

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    the answer that i found id 2(e^t) -1 , why it is wrong ? the answer given is cosht
    XiICijK.png

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 14, 2016 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Did you substitute your answer back into the original differential equation? Did it satisfy this equation and all of the initial conditions?

    If the answer is no, it didn't, then that's why.

    BTW: You should not put one equation in the thread title when the problem covers a completely different equation.
     
  4. Mar 14, 2016 #3

    Mark44

    Staff: Mentor

    Furthermore, the equation should be in the problem statement section, not in the thread title.
     
  5. Mar 14, 2016 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you try your answer and the book's answer in the original equation, you will see both are incorrect. Either that, or you copied the problem incorrectly. In your work, ##\mathcal L(y) \ne s Y(s)##. Also, there is no reason to post this problem as an image instead of just typing it.
     
  6. Mar 16, 2016 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you required to use the Laplace transform? This is a simple second order, linear equation with constant coefficients. It has characteristic equation [itex]r^2- r- 2= (r- 2)(r+ 1)[/itex] and a "specific solution" to the entire equation must be of the form [itex]Ax^2+ Bx+ C[/itex].
     
  7. Mar 16, 2016 #6

    Mark44

    Staff: Mentor

    No. The DE in the title is different from the one in the image that the OP posted. The initial value problem in the posted image is y'' - y = 0, y(0) = 1, y'(0) = 1.
     
  8. Mar 16, 2016 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Which is even easier without using the Laplace transform! The characteristic equation is r^2- 1= (r- 1)(r+ 1)= 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Y" -y' -2y = 4(t^2) , y(0) = 1 , y'(0) =4?
Loading...