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Math Help - Reduction of Order

  1. #1
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    Reduction of Order

    Hi all,
    I'm currently doing a problem based on reduction of order. The question goes:
    Consider the DE L[y] = x(1+3x^2)y''+2y'-6xy = 0. Given that y_{1} = \frac{1}{x} , find a second linearly independent solution.

    I can see that the solution to this equation is in the form y_{2} = u(x)y_{1}. That means y_{2} = u(x)x^{-1}. After rearrange the DE to L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y, I can see that the LDE for u' will be of the form u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0. This simplified to the LDE u''-\frac{6x}{1+3x^2}u'=0.
    To solve for u', I found an integrating factor e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}. Using this integrating factor, solving for u':
    \frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0
    \frac{-6x}{(1+3x^2)^2})u' = C
    Therefore u'=-\frac{C(1+3x^2)^2}{6x}
    When integrating to find u, I'm left with this horrible equation:
    u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D
    I feel this is completely wrong.
    I was wondering if someone could look over my work here and point me in the right direction.
    Thank you in advance.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Silverflow View Post
    Hi all,
    I'm currently doing a problem based on reduction of order. The question goes:
    Consider the DE L[y] = x(1+3x^2)y''+2y'-6xy = 0. Given that y_{1} = \frac{1}{x} , find a second linearly independent solution.

    I can see that the solution to this equation is in the form y_{2} = u(x)y_{1}. That means y_{2} = u(x)x^{-1}. After rearrange the DE to L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y, I can see that the LDE for u' will be of the form u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0. This simplified to the LDE u''-\frac{6x}{1+3x^2}u'=0.
    To solve for u', I found an integrating factor e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}. Using this integrating factor, solving for u':
    \frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0
    \frac{-6x}{(1+3x^2)^2})u' = C
    Therefore u'=-\frac{C(1+3x^2)^2}{6x}
    When integrating to find u, I'm left with this horrible equation:
    u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D
    I feel this is completely wrong.
    I was wondering if someone could look over my work here and point me in the right direction.
    Thank you in advance.
    Process is correct. I don't have time to review the work right now, but what I would suggest is find the general solution via the characteristic equation and compare that to what you have.
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Silverflow View Post
    Hi all,
    I'm currently doing a problem based on reduction of order. The question goes:
    Consider the DE L[y] = x(1+3x^2)y''+2y'-6xy = 0. Given that y_{1} = \frac{1}{x} , find a second linearly independent solution.

    I can see that the solution to this equation is in the form y_{2} = u(x)y_{1}. That means y_{2} = u(x)x^{-1}. After rearrange the DE to L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y, I can see that the LDE for u' will be of the form u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0. This simplified to the LDE u''-\frac{6x}{1+3x^2}u'=0.
    To solve for u', I found an integrating factor e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}. Using this integrating factor, solving for u':
    \frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0
    \frac{-6x}{(1+3x^2)^2})u' = C
    Therefore u'=-\frac{C(1+3x^2)^2}{6x}
    When integrating to find u, I'm left with this horrible equation:
    u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D
    I feel this is completely wrong.
    I was wondering if someone could look over my work here and point me in the right direction.
    Thank you in advance.
    Well, I just finished my Applied Math exam about an hour ago, let's get right back into it!

    Your integrating factor is not correct.

    e^{6 \int \frac{x}{1+3x^2} } = e^{ln(1+3x^2)} = 1+3x^2
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  4. #4
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    Thanks very much, mate.
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  5. #5
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    Actually, the correct integrating factor should be:

    e^p(x)=e^-ln(1+3x^2)=-1/(1+3x^2)

    This is because the p(x) value in this case is negative. Hope this helps!
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