Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential equations questions

  1. Dec 3, 2004 #1
    I was trying to solve a 1st order D.E. and couldn't solve this:

    ln 1/[(y^2)-1] I know the answer is 1/2 ln[(y-1)/(y+1)], but could figure out how this is so.

    Also, I don't understand what method you'd use to get from y = Ae^-4x + Be^-6x to d2y/dx2 +10 dy/dx +24y + 0. I can eliminate the arbitrary constants in simpler D.E.'s, but not this one! Can someone go through a logical method?


    Also, ytan(dy/dx) = (4 + y^2) sec^2(x). How, in this case, would you go about separating the variables?
     
  2. jcsd
  3. Dec 3, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    "Also, I don't understand what method you'd use to get from y = Ae^-4x + Be^-6x to d2y/dx2 +10 dy/dx +24y = 0. I can eliminate the arbitrary constants in simpler D.E.'s, but not this one! Can someone go through a logical method?"
    Eeh, the expression for y(x) is called the GENERAL solution for the DE.
    It SHALL have two arbitrary constants in its expression.
     
  4. Dec 3, 2004 #3

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Zero is a perfectly good constant.

    I have been looking at the first part of this problem but really do not understand the problem. Please give a concise statement of the DE. You may want to investigate our LaTex equation engine.

    As for the second part simply compute the first and second derivative of the given solution, does it satisfy the DE?

    To completly specify A and B you also need 2 initial or boundry conditions. Since those are not specified we can not go any further with the solution.
     
    Last edited: Dec 3, 2004
  5. Apr 8, 2010 #4
    i need help with this problem and please explain the answer:

    Find a linear DE that has x and cosh x as solutions.

    thanks
     
  6. Apr 8, 2010 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In future do not "hijack" other peoples threads to ask new questions. Start your own thread.

    Since cosh(x)= (ex+ e-x)/2, this is the same as asking for a linear d.e. with x, ex, and e-x as solutions. Since x= xe0, any linear d.e. with x as a solution must have characteristice equation with 0 as a double root.

    You need a linear d.e. having a characteristic equation with 0 as a double root and 1 and -1 as roots.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Differential equations questions
  1. Differential Equation (Replies: 3)

  2. Differential Equation (Replies: 12)

Loading...