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Second order differential

  1. Dec 1, 2006 #1
    I need help solving this equation y''+y'+y= sinx

    I know it looks simple but It seems to be getting sticky!! I have been trying to solve it using variation of perimeters,maybe there is a quicker way? If anyone can help please.....:bugeye:
     
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  3. Dec 1, 2006 #2

    selfAdjoint

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    Have you learned how to use Euler's merthod for second order systems with constant coefficients yet?
     
  4. Dec 1, 2006 #3
    ty no, not yet Please explain
     
    Last edited: Dec 1, 2006
  5. Dec 2, 2006 #4
    The form reminds me of a forced oscillator where sin (y) is your forcing function. These aren't too bad. Look at forms along the lines of Ae^yi power as a solution, where i is the imaginary number.

    Note e^i*y= cos(y)+i*sin(y), and since the forcing function you are looking for is sin(y) you will want to look at only the imaginary portion of this equation.


    -----

    There might be faster way to solve this, but I am going to need to know if you have any intial conditions.
     
  6. Dec 2, 2006 #5

    arildno

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    Well, the homogenous part of the solution should be easy to find, to find a particular solution, just insert a trial solution [itex]y_{p}=A\cos(x)+B\sin(x)[/itex] and see what happens.
     
  7. Dec 2, 2006 #6

    HallsofIvy

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    You say you tried variation of parameters so presumably you have already found the general solution to the associated homogenous equation, y"+ Y'+ y= 0. Yes, the solutions to that are a bit complicated:
    [tex]e^{-\frac{x}{2}}\left(C cos(\frac{\sqrt{3}}{2}x)+ D sin(\frac{\sqrt{3}}{2}x)\right)[/tex]
    so that while "variation of parameters" is possible, it involves some complicated algebra.

    Since the function on the right of the equation, sin(x), is one of the solutions one "expects" for linear equations with constant coefficients, do as arildno suggests- use "undetermined coefficients".
     
  8. Dec 2, 2006 #7
    Yes ,That is exactly what i got HallsofIvy, but the integration is very difficult.
     
  9. Dec 2, 2006 #8
    I also tried Yp: A cosx + b cos x , after taking the derivatives they seem to cancel out and leave :A sinx + b cosx = sinx A=1, and B=0 it does not seem to work ? I must be missing something or working it wrong?
     
  10. Dec 2, 2006 #9
    How about this problem did i do this correct ?
    --------------------------------------------------------------------------------


    L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)


    Hence solve the equation y''-3y'+2y=3e^x



    This is how I started to work out this problem.


    y1=e^x This is what i am assuming to be y1 and y2 ?
    y2= Xe^x

    then my W = e^2x

    Yp = 3x^2e^x/2


    Then the general solution Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2

    Did I go about this the wrong way ?

    Should I have plugged the value of L (e^x) and L (Xe^x)
     
    Last edited: Dec 3, 2006
  11. Dec 2, 2006 #10
    OK then lets see what you got !!!!! let me see you break it down for me .
     
  12. Dec 2, 2006 #11

    HallsofIvy

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    Let y= A sin x+ B cos x
    Then y'= A cos x- B sin x
    and y"= -A sin x- B cos x

    y"+ y'+ y= (A- B- A)sin x+ (B+ A- B)cos x= sin x
    or -B sin x+ A cos x = sin x

    You must have B= -1, A= 1.

    The problem did ask you to solve the differential equation, but only after finding L (e^x) and L (Xe^x). Do that first and perhaps you will see how it helps you solve the equation! The solution you give is not correct. In particular, xex is not a solution to the associated homogeneous equation.
     
  13. Dec 3, 2006 #12
    the formula is A cosx + b sinx
    yp: A cos x +b sinx
    y'p: -A sinx +bcosx
    y"p: -A cosx -b sinx


    (A-A+b)cosx +(-A+b-b)sinx= sinx
    B cosx + -Asinx = sinx
    A=-1
    b=0

    Yp: -1 cosx

    y'p : sinx

    y"p : cosx

    now insert them back in gives you y"+y'+y= sinx
    cosx +sinx -cosx =sinx
    sinx = sinx

    Y: e^-x/2[c1cos radical 3/2x + c2 sinx radical 3/2x] -cosx
    ???????
     
  14. Dec 3, 2006 #13
    that should do it.
     
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