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Math Help - Method of Reduction of Order question

  1. #1
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    Method of Reduction of Order question

    Here is what I'm looking at:

    Use the method of reduction of order to find a second solution to

    x^2 y'' + 3xy' + y = 0

    x > 0

    y_1 = x^{-1}



    Here is what I did:


    y'' + \frac{3}{x} y' + x^{-2}y = 0

    p(x) = 3x^{-1}

    y_2 = \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx

    y_2 = \int{\frac{x^{-3}}{x^{-2}}} dx

    y_2 = \int{x^{-1}} dx

    y_2 = ln|x| + c



    Now, when I plug the original ODE into a solver, the general solution comes up with this:

    c_1x^{-1} + c_2 \frac{ln(x)}{x}


    Does that mean that my solution to y_2 is wrong? And if so, where did I make a mistake?
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  2. #2
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Lancet View Post
    Here is what I'm looking at:

    Use the method of reduction of order to find a second solution to

    x^2 y'' + 3xy' + y = 0

    x > 0

    y_1 = x^{-1}



    Here is what I did:


    y'' + \frac{3}{x} y' + x^{-2}y = 0

    p(x) = 3x^{-1}

    y_2 = {\color{red} {y_1}} \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx
    I believe what you're missing is in red above.
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Danny View Post
    I believe what you're missing is in red above.

    Ai! Yes, that would be it, thanks.

    A quick question, though - if I put that in, then the final answer I get is this:


    y_2 = \frac{ln|x| + c}{x}


    Which is decidedly different from what's seen here:

    c_1x^{-1} + c_2 \frac{ln(x)}{x}


    What's causing the discrepancy?
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Lancet View Post
    Ai! Yes, that would be it, thanks.

    A quick question, though - if I put that in, then the final answer I get is this:


    y_2 = \frac{ln|x| + c}{x}


    Which is decidedly different from what's seen here:

    c_1x^{-1} + c_2 \frac{ln(x)}{x}


    What's causing the discrepancy?
    You don't need the second constant of integration with y_2. Note the following

    If y_2 = \dfrac{\ln|x| + k}{x} so

    y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x| + k}{x}.

    If you expand, you can absorb the c_2 k into the c_1.
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Danny View Post
    You don't need the second constant of integration with y_2. Note the following

    If y_2 = \dfrac{\ln|x| + k}{x} so

    y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x| + k}{x}.

    If you expand, you can absorb the c_2 k into the c_1.

    If the k were completely seperable, I could understand that. But because it seperates out into \frac{k}{x}, I was under the impression that we couldn't just do away with that the way we can with a duplicate c by itself.
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Lancet View Post
    If the k were completely seperable, I could understand that. But because it seperates out into \frac{k}{x}, I was under the impression that we couldn't just do away with that the way we can with a duplicate c by itself.
    Post more details on what you mean.
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    Re: Method of Reduction of Order question

    Quote Originally Posted by Danny View Post
    Post more details on what you mean.
    It's like how if you have 2c, c/10, -c, 1/c, e^c, etc... you can take all of those cases and just revert it back to being a simple +c. But as soon as it's attached to a variable, such as cx, you can't revert that back to a simple c.


    If we were looking at:

    y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x|k}{x}


    ...then this could be rewritten as


    y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot k \cdot \dfrac{\ln|x|}{x}

    ...in which case, I can see the two constants being merged. But with what we actually have, seperating out the k would leave it still attached to a variable. And I don't understand how we can merge it with the other constant while it's in that state.
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    Re: Method of Reduction of Order question

    Well, let me provide maybe some more details. What you have is

    \begin{aligned}y &= c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x + k}{x}\\ & = \underbrace{\left(c_1 + c_2 k \right)}_{\text{call this}\; c_1} \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x}{x}\\&=c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln |x|}{x}\end{aligned}
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    Re: Method of Reduction of Order question

    Okay, I get it now! Wow, definitely a bit trickier than usual.

    Thanks for staying with me and helping me understand this!
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    Re: Method of Reduction of Order question

    Glad I could help
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