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Thread: 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

    $\displaystyle x^2 y'' + 3xy' + y = 0$

    $\displaystyle x > 0$

    $\displaystyle y_1 = x^{-1}$



    Here is what I did:


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

    $\displaystyle p(x) = 3x^{-1}$

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

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

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

    $\displaystyle y_2 = ln|x| + c$



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

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


    Does that mean that my solution to $\displaystyle y_2$ is wrong? And if so, where did I make a mistake?
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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

    $\displaystyle x^2 y'' + 3xy' + y = 0$

    $\displaystyle x > 0$

    $\displaystyle y_1 = x^{-1}$



    Here is what I did:


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

    $\displaystyle p(x) = 3x^{-1}$

    $\displaystyle y_2 = {\color{red} {y_1}} \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx$
    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:


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


    Which is decidedly different from what's seen here:

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


    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:


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


    Which is decidedly different from what's seen here:

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


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

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

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

    If you expand, you can absorb the $\displaystyle c_2 k$ into the $\displaystyle c_1$.
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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 $\displaystyle y_2$. Note the following

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

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

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

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

    Quote Originally Posted by Lancet View Post
    If the $\displaystyle k$ were completely seperable, I could understand that. But because it seperates out into $\displaystyle \frac{k}{x}$, I was under the impression that we couldn't just do away with that the way we can with a duplicate $\displaystyle c$ 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 $\displaystyle 2c, c/10, -c, 1/c, e^c$, 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:

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


    ...then this could be rewritten as


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

    ...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

    $\displaystyle \begin{aligned}y &= c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x + k}{x}\\ & = \underbrace{\left(c_1 + c_2 k \right)}_{\text{call this}\; c_1} \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x}{x}\\&=c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln |x|}{x}\end{aligned}$
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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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