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Reduction of order differential eq'n

  1. Nov 14, 2008 #1


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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


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


    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


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


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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


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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


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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


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    Ok here is my attempt (again)




    I then put that into the equation and got


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

    thus getting

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



    [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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