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Which answer is correct? Please check my work

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a second solution y2 for x^2*y"+xy'-y=0; y1=x that isn't a constant multiple of the solution y1.

    2. Relevant equations
    None.

    3. The attempt at a solution
    Here's my work:
    I divided by x^2,
    y"+(1/x)y'-(1/x^2)y=0
    P(x)=1/x and Q(x)=-1/x^2
    Let y(x)=v(x)*x
    y'(x)=v'(x)*x+v(x)
    (1/x)y'(x)=v'(x)+v(x)/x
    y"(x)=v'(x)+v"(x)*x+v'(x)=xv"(x)+2v'(x)
    xv"(x)+2v'(x)+v'(x)+v(x)/x-v(x)*x/(x^2)=0
    xv"(x)+3v'(x)=0
    Let w=v'
    w'=v"
    xw'+3w=0
    w'=-3w/x
    dw/dx=-3w/x
    dw/w=-3/x dx
    integrate
    ln abs(w)=-3ln abs(x)+C
    Don't count C, the constant.
    w=1/x^3
    w=v'=1/x^3
    dv/dx=1/x^2
    don't count c, the constant.
    v=-1/2x^2
    y=v*x
    y=-1/2x^2*x=-1/2x
    But the answer in the book is y2=1/x. I got y=-1/2x, which answer is right?
     
  2. jcsd
  3. Mar 2, 2015 #2

    HallsofIvy

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    Both are correct. If "y" is a solution to a "linear homogenous differential equation", which this is, then "Ay" is also a solution for any constant, A.
     
    Last edited: Mar 3, 2015
  4. Mar 2, 2015 #3

    SammyS

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    If you include a much needed set of parentheses in your solution, y = -1/(2x) , then both are correct.

    Instead of throwing out C, use C = ln(2) .
     
  5. Mar 2, 2015 #4

    LCKurtz

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    What stops you from plugging the answers back in the DE and checking for yourself?
     
  6. Mar 3, 2015 #5

    Mark44

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    I agree. @Math10, at this point in your studies, any problem that you can check, you should check. You have already done the hard work of find the second solution. It's a simple matter to verify that what you have found actually is a solution.
     
  7. Mar 3, 2015 #6
    But when do I use C=ln(2)? After integrating dw/w=-3/x dx? But how did you get C=ln(2)?
     
  8. Mar 3, 2015 #7

    SammyS

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    I assume that you are referring to my response somewhere above.

    Notice that most of the previous respondents used the 'Reply' feature which includes a quote from the post they reference.

    C is an arbitrary constant, Right?

    Suppose you keep it, rather than discarding it.

    You then have ##\displaystyle e^{\,\ln|w|}=e^{\,\ln(|x|^{-3})+C}\ ##.

    So that ##\displaystyle w=\pm e^Cx^{-3}\ ## .

    Carry that on through.

    Then, ##\displaystyle v=\pm\frac{1}{2}e^Cx^{-3}+c_2 ## .

    In your solution, you let c2 = 0. That's fine.

    You can choose C = ln(2), so that ## e^C ## cancels the ##\displaystyle \frac{1}{2}\ ##.

    This together with choosing the - sign from the ± gives the book solution.
     
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