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Math Help - differential equations 2

  1. #1
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    Smile differential equations 2

    just thought that i would post the last 4 differential equations because these are the last 4 i dont know how to do

    1) (d2y/dx2)-3(dy/dx)+2y=0

    2) (d2y/dx2)-3(dy/dx)+2y=e^5x

    3) (d2y/dx2)-3(dy/dx)+2y=e^x

    they all work on eachother so shouldnt be that hard...right?!
    Last edited by emily28; July 19th 2007 at 09:34 AM.
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  2. #2
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    Quote Originally Posted by emily28 View Post
    1) (d2y/dx2)-3(dy/dx)+2y=0
    k^2 - 3k + 2=0
    k=-1,-2
    Thus,
    y=C_1e^x+C_2e^{2x}

    2) (d2y/dx2)-3(dy/dx)+2y=e^5x
    Look for particular of form,
    y = Ae^{5x}

    3) (d2y/dx2)-3(dy/dx)+2y=e^x
    Look for particular of form,
    y=Axe^x

    4) (d2y/dx2)-3(dy/dx)+2y=xe^2x
    Look for particular of form,
    y=Ax^2e^x
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  3. #3
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    Quote:
    2) (d2y/dx2)-3(dy/dx)+2y=e^5x
    Look for particular of form,


    i have used the particular solution of  y=Cxe^{5x}

    so
     y' = Ce^{5x} + 5Cxe^{5x}

    and  y'' = 5Ce^{5x} + 5Ce^{5x} +25 Cxe^{5x}

    which is  Ce^{5x}(25x+10)

    then  Ce^{5x}(25x+10) -3  Ce^{5x} + 5Cxe^{5x} +2Cxe^{5x} = e^{5x}

    giving 10c - 3c = 1
    and c=1/7

    so the general solution is  y= Ae^x + Be^{2x} + 1/7 xe^{5x}


    is this right?
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  4. #4
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    Quote:
    3) (d2y/dx2)-3(dy/dx)+2y=e^x
    Look for particular of form,



    i have used the particular solution of  y=Cxe^{x}

    so
     y' = Ce^{x} + Cxe^{x}

    and  y'' = Ce^{x} + Ce^{x} + Cxe^{x}

    which is  Ce^{x}(x+2)

    then  Ce^{x}(x+2) -3  Ce^{x} + Cxe^{x} +2Cxe^{x} = e^{x}

    giving 2c - 3c = 1
    and c=-1

    so the general solution is  y= Ae^x + Be^{2x} - xe^{x}


    is this right?
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  5. #5
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    are my previous attempts right and does anyone know where i have gone wrong on the last attempt? any advice would be appreciated
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  6. #6
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    please could someone tell me if ive got the previous attempts right? it would really help
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  7. #7
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    Quote Originally Posted by emily28 View Post
    Quote:
    2) (d2y/dx2)-3(dy/dx)+2y=e^5x
    Look for particular of form,


    i have used the particular solution of  y=Cxe^{5x}

    so
     y' = Ce^{5x} + 5Cxe^{5x}

    and  y'' = 5Ce^{5x} + 5Ce^{5x} +25 Cxe^{5x}

    which is  Ce^{5x}(25x+10)

    then  Ce^{5x}(25x+10) -3  Ce^{5x} + 5Cxe^{5x} +2Cxe^{5x} = e^{5x}

    giving 10c - 3c = 1
    and c=1/7

    so the general solution is  y= Ae^x + Be^{2x} + 1/7 xe^{5x}


    is this right?
    this solution is not correct. why didn't you use y = Ae^{5x} as suggested?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emily28 View Post
    Quote:
    3) (d2y/dx2)-3(dy/dx)+2y=e^x
    Look for particular of form,



    i have used the particular solution of  y=Cxe^{x}

    so
     y' = Ce^{x} + Cxe^{x}

    and  y'' = Ce^{x} + Ce^{x} + Cxe^{x}

    which is  Ce^{x}(x+2)

    then  Ce^{x}(x+2) -3  Ce^{x} + Cxe^{x} +2Cxe^{x} = e^{x}

    giving 2c - 3c = 1
    and c=-1

    so the general solution is  y= Ae^x + Be^{2x} - xe^{x}


    is this right?
    this solution is correct.

    good job
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  9. #9
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    cheers its good to know one of them was right, ill have another attempt at 2) and Jhevon do you know if what i have done of my final one is correct?
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  10. #10
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    looking at

    2) (d2y/dx2)-3(dy/dx)+2y=e^5x
    Look for particular of form,


     y' = 5Ce^{5x}

    and  25Ce^{5x}

    then  25Ce^{5x} - 3(5Ce^{5x}) + 2(Ce^{5x}) = e^{5x}

    giving 12c = 1
    and c=1/12

    so the general solution is  y= Ae^x + Be^{2x} + 1/12 e^{5x}


    is this right now ?
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  11. #11
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    Quote Originally Posted by emily28 View Post
    looking at

    2) (d2y/dx2)-3(dy/dx)+2y=e^5x
    Look for particular of form,


     y' = 5Ce^{5x}

    and  25Ce^{5x}

    then  25Ce^{5x} - 3(5Ce^{5x}) + 2(Ce^{5x}) = e^{5x}

    giving 12c = 1
    and c=1/12

    so the general solution is  y= Ae^x + Be^{2x} + 1/12 e^{5x}


    is this right now ?
    Look right to me.
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