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Homework Help: A question about seperable differential equations

  1. Feb 7, 2009 #1
    [itex]\frac{dy}{dx} = \frac{x}{y-1}[/itex]

    You separate it

    [itex] (y-1)\frac{dy}{dx} = x [/itex]

    Some people have moved the 'dx' to the other side and manipulated the dy/dx as if it's just a part of the equation that can be just moved around and that 'dy' and 'dx' are some sort of operators that can be manouvred like this. Others, when they integrate both sides with respect to x show that the 'dx's cancel out on one side leaving with respect to y. And someone said that it can be changed to an integral on the LHS due to the chain rule.

    All of these are understandable but I cannot see them to be very logical and I don't know of any way I can manipulate 'dy's and 'dx's and for them to be of any use so could someone explain why,

    [itex] (y-1)\frac{dy}{dx} = x [/itex],

    [itex]\therefore \int (y-1)\frac{dy}{dx} dx = \int x dx [/itex],

    [itex]\therefore \int (y-1) dy =\int x dx [/itex],

    whether singular dy and dx hold any relevance in solving differential equations and further whether I should attempt to isolate them if so.
  2. jcsd
  3. Feb 7, 2009 #2
    I dont know much reasoning but i know how to do this.
    What i would do is move dy with the y terms and dx with x term. then you could intergre both side and get 1/2(y^2)-y=1/2 x^2
  4. Feb 7, 2009 #3
    Separating the dy and dx is not relevant to solving the problem and is just notational. The theorem being used is that if a(x) = b(x), then the antiderivative of a should be equivalent to the antiderivative of b up to a constant.
    In particular,
    [tex]\int (y-1)\frac{dy}{dx} dx[/tex]
    is evaluated using the change of variable theorem:
    [tex]\int_a^b f(y(x))*y'(x) dx[/tex]
    [tex]= \int_{y(a)}^{y(b)} f(y) dy[/tex]
    Last edited: Feb 7, 2009
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