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Math Help - Quick Diff EQ help

  1. #1
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    Quick Diff EQ help

    here is the problem i've been working on for acouple days and can't get right...

    Given a solution y1=x of the DE x^2y"-xy'+y=0

    Find solution by using the reduction of order method
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chief27 View Post
    here is the problem i've been working on for acouple days and can't get right...

    Given a solution y1=x of the DE x^2y"-xy'+y=0

    Find solution by using the reduction of order method
    I don't have the time at the moment, but read my explanation on reduction of order here. If you're still stuck, please post back.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    I don't have the time at the moment, but read my explanation on reduction of order here. If you're still stuck, please post back.

    I've worked through it using the substitution y=xu then found the first and second derivative of y and plugged them back into the orig. equation..and for the orig equation i divided everything through buy the leading coeff x^2 to modify the equation to y"-1/xy'+1/x^2y

    when i plugged everything into that equation and simplified I got u"+u'+2u

    I then let v=u' to get it to a first order equation
    which gave me

    v'+v+2u and this is where i'm stuck I don't know what to do since there is still a u in the equation
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by chief27 View Post
    here is the problem i've been working on for acouple days and can't get right...

    Given a solution y1=x of the DE x^2y"-xy'+y=0

    Find solution by using the reduction of order method

    y=ux

    y'=u+u'x

    y''=u'+u'+u''x

    x^2(u'+u'+u''x)-x(u+u'x)+ux=0

    2x^2u'+x^3u''-xu-x^2u'+ux=0 \iff x^3u''+x^2u'=0

    Now we need to solve this with the sub v=u' v'=u''

    x^3\frac{dv}{dx}+x^2v=0 \iff \frac{dv}{v}=-\frac{dx}{x}

    \ln|v|=-\ln|x|+\ln(C) \iff \ln|v|=\ln\left( \frac{C}{x}\right) \iff v=\frac{C}{x}

    but v=u' integrating again we get

    u=C\ln|x|+d

    Now if we choose C=1 and d=0 we get

    u=ln|x|

    so the 2nd solution is y=x\ln|x|
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