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Reduction of order differential eq'n

  1. Nov 14, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    Solve : y''+(2/x)y'+y=0 given that y=sinx/x is a solution

    2. Relevant equations



    3. The attempt at a solution
    y=vsinx/x is the other solution

    I worked out

    [tex]y'=\frac{vxcosx-vsinx+v'xsinx}{x^3}[/tex]

    to work out y'' got extremely confusing for me,so I used an online differentiator to do it.
    It gave y'' as

    [tex]\frac{x[2(cosx-sinx)v'+v''xsinx]-v[2xcosx+(x^2-2)sinx]}{x^3}[/tex]

    Now when I substitute it back into the equation, I keep getting the terms for v to not cancel and I get one complex differential equation.
     
  2. jcsd
  3. Nov 15, 2008 #2

    HallsofIvy

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    Your first derivative is incorrect. The denominator should be x2, not x3.
     
  4. Nov 15, 2008 #3

    rock.freak667

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    I will check it over, is there any site where I can input all my differential equations that I need to solve and I can check if I have the correct answer?
     
  5. Nov 15, 2008 #4

    Office_Shredder

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    If you have a possible solution, just plug it into your differential equation and see if it works
     
  6. Nov 15, 2008 #5

    rock.freak667

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    I can do that but I am afraid that I might make a mistake in differentiating the solution and it won't work with the equation and I'll think I did it wrong.
     
  7. Nov 15, 2008 #6

    rock.freak667

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    Ok here is my attempt (again)

    y=vsinx/x

    [tex]y'=(\frac{sinx}{x})v'+v(\frac{cosx}{x}-\frac{sinx}{x^2})[/tex]

    [tex]y''=v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{cosx}{x}-\frac{sinx}{x^2})+v(-\frac{sinx}{x}-\frac{cosx}{x^2}-\frac{cosx}{x^3}+\frac{2sinx}{x^3}[/tex]

    I then put that into the equation and got

    [tex]v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{2cosx}{x})=0[/tex]

    Then let w=v' => w'=v''

    thus getting
    [tex]w'(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+w(\frac{2cosx}{x})=0[/tex]

    [tex]\times x^2[/tex]

    [tex]w'(xsinx+xcosx-sinx)+(2xcosx)w=0[/tex]

    Now

    [tex]\int \frac{1}{w}dw= \int \frac{-2xcosx}{xsinx+xcosx-sinx}dx[/tex]

    and I have no idea how to integrate the term on the right side, I've tried substitutions but it didn't work out well.

    EDIT: my differentiating over complicated the integral...I got it out now.
     
    Last edited: Nov 15, 2008
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