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Homework Help: Find general solution of differential eqn

  1. Jan 31, 2012 #1


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    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]
    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


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    [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


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    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


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    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.
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