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!

Find general solution of differential eqn

  1. Jan 31, 2012 #1

    sharks

    User Avatar
    Gold Member

    The problem statement, all variables and given/known data
    [tex]y''+4y'+3y=\sinh x, \, y(0)=y'(0)=0[/tex]
    The attempt at a solution
    The auxiliary equation is: [itex]m^2+4m+3=0[/itex]
    m= -3 and -1
    The complementary function is: [itex]y_c=Ae^{-3x}+Be^{-x}[/itex]

    I need to find the particular integral next. But i don't know what is the standard inverse D operator result in this case, but i'm going to deduce it from what information i do know.

    [itex]L(D^2)\sinh ax=L(a^2)\sinh ax[/itex]
    From this, i will deduce (maybe i'm wrong) the principle [itex]\frac{1}{L(D^2)}\sinh ax=\frac{1}{L(a^2)}\sinh ax[/itex], where [itex]L(a^2)\not =0[/itex]
    [itex]L(D)=D^2+4D+3[/itex]
    a=1, k=1
    [itex]y_p=\frac{1}{1+4D+3}\sinh x=\frac{1}{4D+4}\sinh x[/itex]
    On multiplying numerator and denominator by (4D-4), [itex]\frac{4D-4}{16D^2-16}\sinh x[/itex]
    And then i applied the same principle again: [itex]\frac{4D-4}{0}\sinh x[/itex]

    And then i got stuck, since the denominator is zero. I must have done something wrong somewhere but i have no idea where.
     
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]Sinh x= (e^x- e^{-x})/2[/itex] so that your solution to the homogeneous equation, [itex]e^{-x}[/itex] is part of the particular solution. Try a particular solution of the form [itex]Ae^x+ Bxe^{-x}[/itex].
     
  4. Jan 31, 2012 #3

    sharks

    User Avatar
    Gold Member

    Unfortunately, i don't understand the method. I need a formula that i can always use whenever i see sinhx but it's unfortunately not listed in my notes.

    For example, in my notes, [itex]\frac{1}{L(D)}ke^{ax}[/itex] gives the generic answer: [itex]\frac{1}{L(a)}ke^{ax}[/itex]

    And also, in my notes, [itex]\frac{1}{L(D)}ksin(ax+b)[/itex] gives the generic answer: [itex]\frac{1}{L(-a^2)}ksin(ax+b)[/itex]

    But there is nothing about [itex]\sinh x[/itex], so i deduced it myself from the above formula:
    [itex]\frac{1}{L(D)}ksinh(ax+b)[/itex] gives the generic answer: [itex]\frac{1}{L(a^2)}ksinh(ax+b)[/itex] but somehow it's wrong as it doesn't solve the problem in post #1.
     
    Last edited: Jan 31, 2012
  5. Jan 31, 2012 #4

    sharks

    User Avatar
    Gold Member

    I finally got it! HallsofIvy, i'm sorry for dismissing your suggestion without looking deeper into it. I converted the R.H.S. to [itex]\frac{e^x-e^{-x}}{2}[/itex] which is the equivalent of [itex]\sinh x[/itex] and the correct answer revealed itself. Thank you for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook