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Integration after implicit differentiation

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the integral of:
    [tex] y + x \frac{dy}{dx}[/tex]

    3. The attempt at a solution
    I know that it is xy, after implicit differentiation.
    However, I cannot get it without prior knowledge of implicit differentiation.
  2. jcsd
  3. Jan 16, 2009 #2


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    Note that you would write the integral of that function as
    [tex]\int (y+ x\frac{dy}{dx})dx= \int ydx+ xdy[/tex]
    That means you are looking for a function F(x,y) so that [itex]dF= F_x dx+ F_y dy= ydx+ xdy[/itex]. If [itex]F_x= y[/itex] then integrating with respect to x, while keeping y constant, F(x,y)= xy+ g(y) where, because we are treating y as a constant, the "constant of integration" may depend on y: g(y). From that, [itex]F_y= x+ g'(y)= x[/itex] which tells us that g'(y)= 0. That means that g really is a constant: g= C so F(x,y)= xy+ C for any constant C.
  4. Jan 16, 2009 #3

    Do you mean this?
    [tex]\int (y+ x\frac{dy}{dx})dx= \int ydx+ \int xdy[/tex]
  5. Jan 16, 2009 #4
    Isn't that what he wrote?
  6. Jan 16, 2009 #5
    I have had an idea that the effect of integral sign ends at the first dx or dy sign. Perhaps, this is not true.
  7. Jan 16, 2009 #6
    Thank you, Hallsofivy!
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