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Math Help - integration

  1. #1
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    integration

    i am having some trouble with the following problem.

    solve:

    dy/dx= (x+1)/y-1), y>1

    here is what is have done so far-

    (y-1) dy/dx=(x+1)

    S[(y-1)dy/dx]dx=S(x+1)dx

    y^2/2 -y=x^2/2+x <-i feel this is where i made a mistake

    Any help would be appreciated.
    thx
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  2. #2
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    Quote Originally Posted by kid funky fried View Post
    i am having some trouble with the following problem.

    solve:

    dy/dx= (x+1)/y-1), y>1

    here is what is have done so far-

    (y-1) dy/dx=(x+1)

    S[(y-1)dy/dx]dx=S(x+1)dx

    y^2/2 -y=x^2/2+x <-i feel this is where i made a mistake Mr F says: Add an arbitrary constant and your answers fine ...... y^2/2 -y=x^2/2+x + C. Personally, I'd multiply through by 2:

    y^2 - 2y = x^2 + 2x + K, where K is just as arbitrary as 2C ......


    Any help would be appreciated.
    thx
    Do you need to make y the subject?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Do you need to make y the subject?
    If so, then

    y^2 - 2y = x^2 + 2x + K \Rightarrow (y - 1)^2 - 1 = x^2 + 2x + K \Rightarrow (y - 1)^2 = x^2 + 2x + A where A = K + 1 is just as arbitrary as K.

    Therefore

    y - 1 = \pm \sqrt{x^2 + 2x + A} \Rightarrow y = \pm \sqrt{x^2 + 2x + A} + 1.

    The value of A and which of the + or - root solutions are kept will depend on the given boundary conditions. Since y > 1 is given ........
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  4. #4
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    well, i did that but then i checked the answer in the back of the book and it reads:

    y=1+ sqrt((x+1)^2+c)
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    If so, then

    y^2 - 2y = x^2 + 2x + K \Rightarrow (y - 1)^2 - 1 = x^2 + 2x + K \Rightarrow (y - 1)^2 = x^2 + 2x + A where A = K + 1 is just as arbitrary as K.

    Therefore

    y - 1 = \pm \sqrt{x^2 + 2x + A} \Rightarrow y = \pm \sqrt{x^2 + 2x + A} + 1.

    The value of A and which of the + or - root solutions are kept will depend on the given boundary conditions. Since y > 1 is given ........
    y = \sqrt{x^2 + 2x + A} + 1 = \sqrt{(x + 1)^2 - 1 + A} + 1 = \sqrt{(x + 1)^2 + B} + 1 = 1 + \sqrt{(x + 1)^2 + B} where B is just as arbitrary as -1 + A.
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  6. #6
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    ok got it thx.
    i'll try another problem.
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