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Help in checking the solution of this separable equation

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data
    It is just an evaluation problem which looks like this dx/dy = x^2 y^2 / 1+x

    2. Relevant equations
    dx/dy = x^2 y^2 / 1 + x

    3. The attempt at a solution
    What i did is cross multiply to get this equation y^2 dy = x^2 / 1+x dx then next line ∫y^2 dy = ∫x^2/1+x dx

    y^3/3 = ∫dx + ∫1/x dx after simplifying i get y^3=3x + 3 ln x + C but im not sure if this is the right answer.
     
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  3. Jul 8, 2015 #2

    HallsofIvy

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    First, I presume that you really mean dx/dy= x^2y^2/(1+ x). What you wrote, dx/dy= x^2y^2/1+ x, is the same as dx/dy = x^2y^2+ x.

    More importantly you have separated the x and y incorrectly. The left side is y^2 dy but the right sides should be [(1+ x)/x]dx. You have the fraction inverted.
     
  4. Jul 9, 2015 #3
    oh, so when i integrate [1+x / x] dx would it look like ∫1/x + ∫x dx? or something else?
     
  5. Jul 9, 2015 #4

    LCKurtz

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    1 + x/x = 1+1 =2.
     
  6. Jul 9, 2015 #5

    Mark44

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    Or did you intend the above to mean (1 + x)/x? If that's what you intended, it does absolutely no good to put brackets around the entire expression.

    In post #1 you wrote what I've quoted below. For that, you need parentheses around the denominator, as x2y2/(1 + x).

     
  7. Jul 10, 2015 #6
    Oh sorry about that, what i intend to do is to integrate the entire expression. As it is the right hand side of the equation, no problem with y^2 dy but the right side looks a bit confusing. In the expression (1 + x) / x dx as was suggested is what i intend to integrate, so is this the right integration expression, ∫1 / x dx + ∫ dx? if so then ln x + x + C should be sufficient for the RHS, is it not?
     
  8. Jul 10, 2015 #7

    Mark44

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    Yes, you can split ##\int \frac{1 + x}{x} \ dx## into ##\int \frac {dx} x + \int 1 \ dx##.
     
  9. Jul 11, 2015 #8
    so the answer y^3 = 3x + 3 ln x + C should be correct? ok i get it now thanks for the time.
     
  10. Jul 12, 2015 #9

    Mark44

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    No @enggM, this is not correct. The work you did before was incorrect, and my earlier response was based on that work. I think you need to start from the beginning.

    $$\frac{dx}{dy} = \frac{x^2y^2}{1 + x}$$

    If you multiply both sides by 1 + x, then divide both sides by ##x^2##, and finally, multiply both sides by dy, the equation will be separated. What do you get when you do this?
     
  11. Jul 19, 2015 #10
    @Mark44 i would get (1+x / x^2 ) dx = y^2 dy so integrating the both sides y^3 / 3 = 1 / x + ln x so the final form would then be y^3 = 3/x + 3 ln x + C? no?
     
  12. Jul 19, 2015 #11

    Ray Vickson

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    From
    [tex] y^2 dy = \left( 1 + \frac{x}{x^2} \right) dx [/tex]
    you will get
    [tex] \frac{1}{3} y^3 = x + \ln (x) + C[/tex]
    From
    [tex] y^2 dy = \frac{1 + x}{x^2} dx [/tex]
    you will get
    [tex] \frac{1}{3} y^3 = -\frac{1}{x} + \ln (x) + C[/tex]

    Which do you mean? Why are you still refusing to use parentheses? Do you not see their importance?
     
  13. Jul 19, 2015 #12

    Mark44

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    In writing (1 + x/x2) above, you are using parentheses, but aren't using them correctly (as Ray also points out). If you have a fraction where either the numerator or denominator (or both) has multiple terms, you need parentheses around the entire numerator or denominator, not around the overall fraction.

    If you mean ##\frac{1 + x}{x^2}##, write it in text as (1 + x)/x^2, NOT as (1+x / x^2 ). What you wrote means ##1 + \frac x {x^2} = 1 + \frac 1 x##.
     
  14. Jul 28, 2015 #13
    sorry about the confusion because sometimes when i solve something like this in paper i sometimes leave out the parenthesis. so the final form would be y^3 = -3/x + 3 ln x + C? by the way how did it become a negative? just an additional question.
     
  15. Jul 28, 2015 #14

    SammyS

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    What is ##\displaystyle \int\frac{1}{x^2}dx\ ?##
     
  16. Jul 29, 2015 #15

    Mentallic

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    On paper you can freely write the fraction as it's normally portrayed, however, if you mean that you do sometimes write everything on a single line and still don't use parentheses, then you need to break out of that habit. Parentheses are crucial.
     
  17. Aug 5, 2015 #16
    @SammyS oh i see i remember now it should be u^-2+1 / -2 +1.
    @Mentallic ok thanks...
    thanks for all of your help in checking for the solution and answer to this problem helped me understand it much better.
     
  18. Aug 5, 2015 #17

    LCKurtz

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    Apparently you haven't been paying attention to anything people in this thread have told you about the importance of using parentheses.
     
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