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Math Help - Substitution question

  1. #1
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    Substitution question

    I've been stuck on this problem all day and can't seem get my solution to match the one in my textbook.

    Use the appropriate method of substitution to solve the equation:

    xdx + (y-2x)dy = 0

    The xdx part is giving me some trouble. I tried using u = y/x but I only got to this point:

    \int dx/x = \int 1/(u-1)du

    The answer from the book:

    (x-y)ln|x-y| = y + c(x-y)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mdmac View Post
    I've been stuck on this problem all day and can't seem get my solution to match the one in my textbook.

    Use the appropriate method of substitution to solve the equation:

    xdx + (y-2x)dy = 0

    The xdx part is giving me some trouble. I tried using u = y/x but I only got to this point:

    \int dx/x = \int 1/(u-1)du

    The answer from the book:

    (x-y)ln|x-y| = {\color{red}y}\, + c(x-y)
    First, divide the equation through by x to get \,dx +\left(\frac{y}{x}-2\right)\,dy=0

    Now applying the substitution y=ux\implies \,dy=u\,dx+x\,du, we see that the DE becomes

    \,dx+\left(u-2\right)\left(u\,dx+x\,du\right)=0\implies \,dx+u^2\,dx+ux\,du-2u\,dx-2x\,du=0 \implies -\left(u^2-2u+1\right)\,dx=\left(u-2\right)x\,du

    Now separating the variables, we have -\frac{\,dx}{x}=\frac{u-2}{\left(u-1\right)^2}\,du

    Now we integrate. The LHS becomes -\ln x + C.

    Now we need to do some manipulation on the RHS.

    Note that \frac{u-2}{\left(u-1\right)^2}=\frac{u-1}{\left(u-1\right)^2}-\frac{1}{\left(u-1\right)^2}=\frac{1}{u-1}-\frac{1}{\left(u-1\right)^2}. Now when we integrate wrt u, we have \ln\!\left|u-1\right|+\frac{1}{u-1}+C.

    So it turns out that we have -\ln x + C=\ln\!\left|u-1\right|+\frac{1}{u-1}+C\implies -\ln x= \ln\!\left|u-1\right|+\frac{1}{u-1}+K.

    Backsubstituting u=\frac{y}{x}, we have -\ln x=\ln\!\left|\frac{y}{x}-1\right|+\frac{1}{\frac{y}{x}-1}+K\implies -\ln x=\ln\!\left|y-x\right|-\ln x+\frac{x}{y-x}+K

    Continuing with the simplification, we have \frac{x}{x-y}+K=\ln\!\left|x-y\right|\implies {\color{red}x}\,+K\left(x-y\right)=\left(x-y\right)\ln\!\left|x-y\right|

    Hmm...

    My solution and the books solution disagree...due to one variable...XD

    Are you sure you copied it down correctly?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    First, divide the equation through by x to get \,dx +\left(\frac{y}{x}-2\right)\,dy=0

    Now applying the substitution y=ux\implies \,dy=u\,dx+x\,du, we see that the DE becomes

    \,dx+\left(u-2\right)\left(u\,dx+x\,du\right)=0\implies \,dx+u^2\,dx+ux\,du-2u\,dx-2x\,du=0 \implies -\left(u^2-2u+1\right)\,dx=\left(u-2\right)x\,du

    Now separating the variables, we have -\frac{\,dx}{x}=\frac{u-2}{\left(u-1\right)^2}\,du

    Now we integrate. The LHS becomes -\ln x + C.

    Now we need to do some manipulation on the RHS.

    Note that \frac{u-2}{\left(u-1\right)^2}=\frac{u-1}{\left(u-1\right)^2}-\frac{1}{\left(u-1\right)^2}=\frac{1}{u-1}-\frac{1}{\left(u-1\right)^2}. Now when we integrate wrt u, we have \ln\!\left|u-1\right|+\frac{1}{u-1}+C.

    So it turns out that we have -\ln x + C=\ln\!\left|u-1\right|+\frac{1}{u-1}+C\implies -\ln x= \ln\!\left|u-1\right|+\frac{1}{u-1}+K.

    Backsubstituting u=\frac{y}{x}, we have -\ln x=\ln\!\left|\frac{y}{x}-1\right|+\frac{1}{\frac{y}{x}-1}+K\implies -\ln x=\ln\!\left|y-x\right|-\ln x+\frac{x}{y-x}+K

    Continuing with the simplification, we have \frac{x}{x-y}+K=\ln\!\left|x-y\right|\implies {\color{red}x}\,+K\left(x-y\right)=\left(x-y\right)\ln\!\left|x-y\right|

    Hmm...

    My solution and the books solution disagree...due to one variable...XD

    Are you sure you copied it down correctly?
    I just doubled checked and the book shows a y rather than an x. Your explanation so far has been extremely helpful though.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mdmac View Post
    I just doubled checked and the book shows a y rather than an x. Your explanation so far has been extremely helpful though.
    I checked my solution through another means, and it should be x, not y. So maybe its a typo in your book...
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    I checked my solution through another means, and it should be x, not y. So maybe its a typo in your book...
    I really does seem that way. Man, what a lousy deal. Thank you for your help though, I was pulling my hair out!
    Last edited by mr fantastic; June 23rd 2009 at 02:38 AM.
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