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Reduction of order (2nd order linear ODE homogeneous ODE)

  1. Aug 31, 2007 #1
    Reduce to the 2nd order and solve the eq.

    (1-x^2)y'' - 2xy' + 2y = 0 (we know that y(1) = 0)



    I don't know what to do here, except that I'm supposed to find y(2). I know the formulas

    y(2) = y(1) Int(U) dx

    where

    U = (e^(-Int(p(x)dx)) / (y(1))^2

    Can I use this?
     
    Last edited: Aug 31, 2007
  2. jcsd
  3. Aug 31, 2007 #2
    Updated
     
  4. Aug 31, 2007 #3

    rock.freak667

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    To reduce the order, I believe one soltion [tex]y_1[/tex] must be known and then the other solution is [tex]y_2 =vy_1[/tex] do you know the at least one solution?
     
  5. Aug 31, 2007 #4
    [tex]y_1[/tex] = x

    That's what I meant by Y(1)=0. I don't know how to write these math typings.
     
    Last edited: Aug 31, 2007
  6. Aug 31, 2007 #5

    rock.freak667

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    If y(1)=0 then y(2) = y(1) Int(U) dx then wouldn't y(2) also become 0?
     
  7. Aug 31, 2007 #6
    Haha, my mistake. y(1)=x, not 0!
     
  8. Aug 31, 2007 #7

    rock.freak667

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    then it becomes simpler then [tex]y_1=x[/tex] then [tex] y_2=vx[/tex]
    then y=vx => [tex]\frac{dy}{dx}=x\frac{dv}{dx} + v[/tex]
    and [tex]\frac{d^2y}{dx^2}=2\frac{dv}{dx} +x\frac{d^2v}{dx^2}[/tex]
    sub those into the equation and you'll get[tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex]

    then let any variable you want be [tex]\frac{dv}{dx}[/tex]...say [tex]w=\frac{dv}{dx}[/tex] and then [tex]\frac{dw}{dx}=\frac{d^2v}{dx^2}[/tex] and then you solve from there

    i.e [tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex] becomes

    [tex](x-x^3)\frac{dw}{dx} + (2-2x)w=0[/tex]

    and what you did there was reduce the 2nd order linear ODE homogeneous eq'n to a first order diff eq'n
     
  9. Aug 31, 2007 #8
    I don't understand why I'm supposed to get: [tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex]

    What I get is: [tex](x-x^3)\frac{d^2v}{dx^2} + (2-4x^2)\frac{dv}{dx} - 4x*v=0[/tex]
     
  10. Aug 31, 2007 #9

    rock.freak667

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    ah sorry i forgot to add a term

    But this is how the algebra is supposed to be

    [tex](1-x^2)(2\frac{dv}{dx}+x\frac{d^2v}{dx^2})-2x(\frac{dv}{dx}+v)+2(vx)=0[/tex]

    [tex]2\frac{dv}{dx}-2x^2\frac{dv}{dx}+x\frac{d^2v}{dx^2}-x^3\frac{d^2v}{dx^2}-2x^2\frac{dv}{dx}-2vx+2vx=0[/tex]

    finally giving:

    [tex](x-x^3)\frac{d^2v}{dx^2}+(2-4x)\frac{dv}{dx} =0[/tex]

    then sub [tex]w=\frac{dv}{dx}[/tex]
     
    Last edited: Aug 31, 2007
  11. Sep 1, 2007 #10
    OK, I guess you mean

    [tex](x-x^3)\frac{d^2v}{dx^2}+(2-4x^2)\frac{dv}{dx} =0[/tex]

    Now then, how do I fine the 2nd solution?
     
  12. Sep 1, 2007 #11

    rock.freak667

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    Let [tex]w= \frac{dv}{dx}[/tex]
    [tex]\frac{dw}{dx} = \frac{d^2v}{dx^2}[/tex]

    then sub those in your equation and you will get a variables are separable type of diff eq'n

    [tex](x-x^3)\frac{dw}{dx} +(2-4x)w = 0 [/tex]


    [tex](x-x^3)\frac{dw}{dx}= -(2-4x)w [/tex]

    [tex]\frac{1}{w}\frac{dw}{dx} = \frac{-(2-4x)}{(x-x^3)}[/tex]

    integrate both sides w.r.t x now
     
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