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Seperation of Variables

  1. Jun 23, 2008 #1
    The problem statement, all variables and given/known data

    [tex](x-1)y'=6y[/tex]

    Could someone please help explain to me how to do this problem.

    Here's my attempt....

    [tex](x-1)y'=6y[/tex]
    [tex](x-1)dy=6y[/tex]
    [tex](x-1)=6y/dy[/tex]
    [tex]\int (x-1)dx=\int 6y/dy[/tex]
    [tex]((x^2,2)-x)= 3y^2+C[/tex]

    Then I try to get it to equal y and it comes out nothing like the answer. Am I even doing these steps correct?
     

    Attached Files:

  2. jcsd
  3. Jun 23, 2008 #2

    Dick

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    y'=dy/dx. Put that in and separate. Don't put dy or dx into the denominator. It's supposed to be in the numerator to integrate.
     
  4. Jun 23, 2008 #3
    ok so it would be like this....

    [tex](x-1)dy/dx=6y[/tex]
     
  5. Jun 23, 2008 #4

    Dick

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    Yes. Now separate.
     
  6. Jun 23, 2008 #5
    ok now heres what i have...

    [tex](x-1)dy/dx=6y[/tex]
    [tex](x-1)dy=6ydx[/tex]
    [tex]dy=6ydx/(x-1)[/tex]
    [tex]dy/6y=dx/(x-1)[/tex]
    [tex]
    \int dy/6y=\int dx/(x-1)
    [/tex]

    is that correct?
     
  7. Jun 23, 2008 #6

    Dick

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    That's it. Good job.
     
  8. Jun 23, 2008 #7
    ok but now I need to solve for y, so are these the right answers for the integral?

    [tex]

    \int dy/6y=\int dx/(x-1)

    [/tex]
    [tex] (y^2/12)=ln(x-1)+c [/tex]
     
  9. Jun 23, 2008 #8

    Dick

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    No, y is in the denominator on the left side. It should give you another log type integral.
     
  10. Jun 23, 2008 #9
    [tex] ln(y)/6 [/tex] is that right? and is the other side correct?
     
  11. Jun 23, 2008 #10

    Dick

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    Yes. Now both sides are correct. Can you solve for y now?
     
  12. Jun 23, 2008 #11
    [tex] ln(y)/6 = ln(x-1) + C [/tex]

    Now how do I solve for y? I am not sure what to do with the ln's.
     
  13. Jun 23, 2008 #12
    **edit**

    *erased*
     
  14. Jun 23, 2008 #13

    Dick

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    Exponentiate both sides. exp(ln(A))=A. I'd move the 6 over to the x side first.
     
  15. Jun 23, 2008 #14
    so it would be..

    e^(ln(y))=6e^(ln(x-1)) + C
     
  16. Jun 23, 2008 #15

    Dick

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    Noo. e^(6*ln(x-1)) is not equal to 6*e^(ln(x-1)). And e^(A+B) is not equal to e^(A)+e^(B). Be careful! Review log rules.
     
  17. Jun 23, 2008 #16
    Ohhh my mistake. e^(ln(y))=e^6ln(x-1)

    y=(x-1)^6

    is that correct now?
     
  18. Jun 23, 2008 #17

    Dick

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    Almost, where did the constant go?
     
  19. Jun 26, 2008 #18
    y=C(x-1)^6
     
  20. Jun 26, 2008 #19

    Dick

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    That's it.
     
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