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Math Help - Solving Differential Equation

  1. #1
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    Solving Differential Equation

    I'm not getting the correct answer here and I'm sure i've got something screwed up.
    given: r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

    This is my workout:
    r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

    -try y=r^n
    y'=nr^(n-1)
    y''=n(n-1)r^(n-2)
    - plugin
    r^2(n(n-1))+6(nr^(n-1))+6(r^n)=0
    n(n-1)+6n+6=0
    (n^2)-n+6n+6=0
    (n^2)+5n+6=0
    -where
    n=3, n=2
    -general equation
    y(r)=C1r^(3n) + C2r^(2n)
    -take derivative of general equation which gives
    y'(r)=3C1(ln(r)+1)r^(3r) + 2C2(ln(r)+1)r^(2r)

    -plugin value y(2)=0 into general equ and solve for C1 and C2
    0=C1(2)^(3*2) + C2(2)^(2*2)
    C1=-C2/4 C2=-4C1
    -Plugin value y'(2)=1 into the derivative
    1=3C1(ln(2)+1)2^(3*2) + 2C2(ln(2)+1)2^(2*2)
    C1=-(2C2-1)/192*(ln(2)+1) C2=(1/2)-96C1*(ln(2)+1)

    So what do I do to finish out the problem? From the book the answer is y=(11/3e)r^2 - ((11e^(2))/3)r^-1
    Thank you for your help in advance!
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  2. #2
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    Re: Solving Differential Equation

    Quote Originally Posted by nivek0078 View Post
    I'm not getting the correct answer here and I'm sure i've got something screwed up.
    given: r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

    This is my workout:
    r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

    -try y=r^n
    y'=nr^(n-1)
    y''=n(n-1)r^(n-2)
    - plugin
    r^2(n(n-1))+6(nr^(n-1))+6(r^n)=0
    n(n-1)+6n+6=0
    (n^2)-n+6n+6=0
    (n^2)+5n+6=0
    -where
    n=3, n=2
    -general equation
    y(r)=C1r^(3n) + C2r^(2n)
    -take derivative of general equation which gives
    y'(r)=3C1(ln(r)+1)r^(3r) + 2C2(ln(r)+1)r^(2r)

    -plugin value y(2)=0 into general equ and solve for C1 and C2
    0=C1(2)^(3*2) + C2(2)^(2*2)
    C1=-C2/4 C2=-4C1
    -Plugin value y'(2)=1 into the derivative
    1=3C1(ln(2)+1)2^(3*2) + 2C2(ln(2)+1)2^(2*2)
    C1=-(2C2-1)/192*(ln(2)+1) C2=(1/2)-96C1*(ln(2)+1)

    So what do I do to finish out the problem? From the book the answer is y=(11/3e)r^2 - ((11e^(2))/3)r^-1
    Thank you for your help in advance!
    Everything is good until right after this line

    n^2+5n+6

    is correct but,

    (n+2)(n+3)=0 \iff n=-2 \quad n=-3

    So the solution should look like

    y(r)=\frac{c_1}{r^2}+\frac{c_2}{r^3}
    Thanks from nivek0078
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  3. #3
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    Re: Solving Differential Equation

    How would you solve this problem?

    Theta squared d Theta = sin(t + 0.2) dt ?
    Last edited by Jason76; October 19th 2012 at 06:46 AM.
    Thanks from nivek0078
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