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!

Differential Calculus variaton of parameters question

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data
    The equation that has to be solved:
    y'' - y' - 2y = 2e^(-t)


    The problem I am having is that I don't understand why they equatate that part with the derivatives of the u parameters to 0. (see image)
    2dwafs2.png

    Here they first find the characteristic equation and write down the general solution. They then replace the constants with the parameter "u" and take the derivate.

    As you can see, they just say that the derivate part of the u parameter is equal to 0. But why? How? Where did that come from? I can't find it anywhere in my book.

    It's probably a facepalm answer but I would really appreciate it
     
  2. jcsd
  3. Jan 7, 2013 #2

    jedishrfu

    Staff: Mentor

    Isnt the derivative of a constant zero?
     
  4. Jan 7, 2013 #3
    yes but for some reason in the examples and solutions they even take the second derivative of u and they leave it in the equation. It's just that those 2 "parts" are equal to 0 for some reason that is unbeknownst to me and my buds here in the library..
     
  5. Jan 7, 2013 #4

    jedishrfu

    Staff: Mentor

    they may be showing you generalized steps that you should keep in mind for the real world and then reducing them since the u1(t)=constant hence u1'=0 and u1''=0 ...
     
  6. Jan 7, 2013 #5
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differential Calculus variaton of parameters question
Loading...