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Homework Help: First Order ODE

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    So First I checked if both equations were exact. I took the derivative of 3xy^2+4y and also derivative of the other and they were both equal so the equation is exact.

    I took the 3xy^2+4y and integrated it with respect to x. And then I differentiated it to y.

    I got 3x^2y+4x out of it. The problem is that it is exactly the same as the one from the originally equation so I did not ending up finding C.

    Did I do something wrong?
    Last edited by a moderator: Feb 2, 2015
  2. jcsd
  3. Feb 2, 2015 #2


    Staff: Mentor

    I agree that the equation is exact. Where you went wrong is in your integration.
    $$\int (3xy^2 + 4y)dx \neq 3x^2y + 4x$$
    When you integrate the above, treat y as if it were a constant.
    Do a similar integration, with respect to y, for the other part.

    Your textbook should have some examples of how this works.
  4. Feb 2, 2015 #3
    No after I integrated, I got 3/2x^2y^2+4xy

    Then differentiate and got 3x^2y+4x
  5. Feb 2, 2015 #4
    The function you are looking for is the one after you integrated. If you take the derivative and you get back to that function, you do not need to account for any extra functions. It looks correct.

    ∫(Fx)dx = F(x,y) + F(y) + C ... (1)
    ∫(Fy)dy = F(x,y) + F(x) + C ... (2)

    In this case, if you integrate (1), take the derivative with respect to y, and compare it to Fy that's given, you should be able to determine any term that was missing. In this case, there are no terms to account for
  6. Feb 3, 2015 #5
    So C = 0?
  7. Feb 3, 2015 #6


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    You miss the integration constant, which can be function of y. So the integral is 3/2x^2y^2+4xy+C(y).
    Differentiate it with respect y. It is 3x^2y+4x+C'. It must be equal to 3x^2y+4x, so C' = 0. It means that C is a ????
  8. Feb 3, 2015 #7
    ∫(Fx)dx = Φ(x,y) + F(y) + C ... (1)
    ∫(Fy)dy = Φ(x,y) + F(x) + C ... (2)
    F(y) and F(x) are zero. C is just C, the constant of integration (a number and not a function of x or y). There is a distinction between functions of y or x that arise from integration w.r.t the other variable, and the standard integration constant that arises from any integration. The former is zero, while the latter is to be left undetermined.

    The solution to an exact equation is written in the form:
    Φ(x,y) + f(x) + f(y) = C. In this case, f(x) and f(y) are just zero, so:
    Φ(x,y) = C should be the answer
    Last edited: Feb 3, 2015
  9. Feb 3, 2015 #8


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    Has a speck of dust surrounded my brain like sometimes, or are you meant to notice that the original factorises such that it reduces to a distinctly easier problem than it is looking?
    Last edited: Feb 3, 2015
  10. Feb 3, 2015 #9

  11. Feb 3, 2015 #10


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    Too clever and unnecessary IMHO.
  12. Feb 4, 2015 #11


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    No one? Won't the OP look at the two expressions, see that they obviously have a factor, that leads to solving the problem in one or two lines?
  13. Feb 9, 2015 #12


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    Well my hint has been up for several days now.

    Isn't this just

    (3xy + 4)(y dx + x dy) = 0

    so that all you have to solve is

    dy/y = - dx/x

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