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Finding particular solution yp of given equation

  1. Sep 6, 2011 #1
    y''+2y'-3y=1+xex

    ive tried yp=Axex+B

    and yp=Ax2ex+Bx

    and both dont work. assuming im doing it correctly.

    im given the answer as: 1/3+1/16(2x2-x)ex

    suggestions?

    thanks
     
  2. jcsd
  3. Sep 6, 2011 #2

    HallsofIvy

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    [itex]e^x[/itex] is a solution to the associated homogeneous equation so if the function on the right were just [itex]e^x[/itex] you would have to try [itex]xe^x[/itex]. Because you want [itex]xe^x[/itex], try [itex]Ax^2e^x[/itex] instead.
     
  4. Sep 6, 2011 #3

    LCKurtz

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    The proper choice is to try yp = Axex+Bx2ex.
     
  5. Sep 6, 2011 #4
    Thanks guys, LC that makes sense now I see it, I was varying the order of the ex^x but not the 1. Which of course is silly.
     
  6. Sep 7, 2011 #5
    heyy...

    i tried yp=Axex+Bx2ex and got

    after subbing back into y''+2y'-3y

    result was A(4ex)+B(8xex+2ex)=1+xex

    looking at that i saw i could cancel out constant A by making it Aex

    using Aex+Bxxex

    after subbing in i got B(8xex+2ex)=1+xex


    how can i possibly solve for A or B??
     
  7. Sep 7, 2011 #6

    HallsofIvy

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    [itex]y= (Ax+ Bx^2)e^x[/itex]
    [itex]y'= (A+ 2Bx)e^x+ (Ax+ Bx^2)e^x= (A+ (2B+ A)x+ Bx^2)e^x[/itex]
    [itex]y''= (2B+ A+ 2B)e^x+ (A+ (2B+ A)x+ Bx^2)e^x= (6B+ 2A+ (2B+ A)x+ Bx^2)e^x[/itex]
    [itex]y''+ 2y'- 3y= [(6B+ 4A)+ 4Bx]e^x= xe^x[/itex]

    Solve for A and B.
     
  8. Sep 7, 2011 #7
    i dont know how to do the quotes thing so ill copy and paste.

    " y′′+2y′−3y=[(6B+4A)+4Bx]ex=xex "


    but from the original question... y′′+2y′−3y=xex +1

    whered the +1 go?
     
  9. Sep 7, 2011 #8

    HallsofIvy

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    Because the equation is linear, you can do that separately.

    Let y= B. Then y''= y'= 0 so the equation becomes ...
     
  10. Sep 7, 2011 #9
    ohhh. I missed that :/

    hang around please, gimme 10min to post the way i did it. its slightly different.
     
  11. Sep 7, 2011 #10
    y= Axex+bx2ex

    y'= A(xex+ex) + B(2xex+x2ex)

    y''= A(xex+ex+ex) + B(2xex+2ex+2xex+x2ex)

    subbing in

    y''+2y'-3y = [A(xex+ex+ex) + B(2xex+2ex+2xex+x2ex)] + 2[A(xex+ex) + B(2xex+x2ex)] - 3[Axex+bx2ex]

    = A[xex+2ex+2xex+2ex-3xex] + B[4xex+2ex+x2ex+4xex+2x2ex-3x2ex

    = A[4ex] + B[8xex+2ex]

    the only thing that differs from your solution is the B parts

    it seems like one of my 2xex should only be 2ex so i get 6 and 4 instead of 8 and 2....
     
  12. Sep 7, 2011 #11

    HallsofIvy

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    I had before, [itex]y′′+2y′−3y=[(6B+4A)+4Bx]e^x=xe^x[/itex]
    so you must have 4B= 1 and 6B+ 4A= 0.
     
  13. Sep 7, 2011 #12
    yeah i did that part. there just an error in my jumble up there. was hoping you'd see it. but nevermind. thanks for all your help though, much appreciated! :)
     
  14. Sep 7, 2011 #13
    yeah.... i was right...

    to get the correct answer the euqations for A and B should be

    8B=1 and 2B+4A=0

    gives B = 1/8 and A = -1/16
     
  15. Sep 7, 2011 #14
    about that +1 that got left alone... you said we can do that seperately...

    y''+2y'-3y=1+xex

    taking out all differentiable functions leaves 3y=-1 so y= -1/3 + yp ??
     
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