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Math Help - Reducible Second Order Equation

  1. #1
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    Question Reducible Second Order Equation

    Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

    (x^2)y'' + 3xy'=2

    This is what I have but came up with the wrong answer
    can someone explain what I need to do (Thanks)

    y is missing
    Let p= y'= dy/dx then let p'=y''=dp/dx

    (x^2) dp/dx +3xp= 2

    dp/dx + (3/x)p = (2/x^2)

    Now i used the linear first-order equation and found

    Dx + (3x^2)p = 2x

    I then got

    (3x^2)p = x^2 + C

    Dividing by (3x^2) I got

    p= (1/3) + C/(3x^2)

    This is not the right answer, the right answer is ln(x) + Ax^-2 + B
    I do not have a clue how they got this can someone please help me???
    Thanks
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

    (x^2)y'' + 3xy'=2
    You need to remember that you solved for p.
    The problem asks to solve for y.
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  3. #3
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    Quote Originally Posted by mandy123 View Post
    Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

    (x^2)y'' + 3xy'=2

    This is what I have but came up with the wrong answer
    can someone explain what I need to do (Thanks)

    y is missing
    Let p= y'= dy/dx then let p'=y''=dp/dx

    (x^2) dp/dx +3xp= 2

    dp/dx + (3/x)p = (2/x^2)

    Now i used the linear first-order equation and found

    Dx + (3x^2)p = 2x

    I then got

    (3x^2)p = x^2 + C

    Dividing by (3x^2) I got

    p= (1/3) + C/(3x^2)

    This is not the right answer, the right answer is ln(x) + Ax^-2 + B
    I do not have a clue how they got this can someone please help me???
    Thanks
    So far you have only found  p . Remember that  p = \frac{dy}{dx} . What must you do to  \frac{dy}{dx} to get y?

    But you solved incorrectly for p anyway.

     \frac{dp}{dx} +\frac{3}{x}p = \frac{2}{x^2}

     \phi = e^{\int \frac{3}{x}dx}  = e^{3\ln|x|} = e^{\ln(x^3)} = x^3

     x^3 p = \int x^3 \times \frac{2}{x^2}dx

     x^3 p = \int 2xdx

     x^3 p = x^2+A

     p = \frac{1}{x}+\frac{A}{x^3}

     y= \int \frac{1}{x}+\frac{A}{x^3}dx

     y= \ln|x|-\frac{A}{2x^2}+B
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  4. #4
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    Thanks for letting me know I solved for p wrong, I realized my mistake when you said something.
    I agree with this answer, but why does the answer key show the answer is ln(x) + Ax^-2 + B
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  5. #5
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    Quote Originally Posted by mandy123 View Post


    Thanks for letting me know I solved for p wrong, I realized my mistake when you said something.
    I agree with this answer, but why does the answer key show the answer is ln(x) + Ax^-2 + B
    Well taking an arbitrary constant, multiplying it by negative 1 and dividing it by 2 still gives you a an arbitrary constant.

    So think of it as:

     y = \ln|x| - \frac{A^*}{2x^2} + B

     = \ln|x| + Ax^{-2} + B

    Where  A = -\frac{A}{2}

    Yes?
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  6. #6
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    Ok, sounds good, thank you so much
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