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Exact Equation, reviewing for an exam, did I do this one right?

  1. Feb 26, 2006 #1
    Hello everyone, its exam time so i'm just reviewing all the different techniques for Diff EQ and i hit an exact diff EQ problem and i was just wondering if someone can check to see if i did it right. HEre it is:
    Directions are find a relationships between x and y of the form f(x,y) = c.
    [​IMG]

    Thanks! :biggrin:
     
  2. jcsd
  3. Feb 26, 2006 #2

    quasar987

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    Yes, except for the line [itex]\Psi(x,y) = \int Mdx[/itex] which should read [itex]\Psi(x,y) = \int Mdx + h(y)[/itex], it's good. But in an exam, you might want to explain what and why you do these things more.
     
  4. Feb 26, 2006 #3
    Awesome thanks quasar
     
  5. Feb 27, 2006 #4

    HallsofIvy

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    On the contrary, a lot of what you have written is non-sense!

    You are correct that, with [itex]\Phi_x[/itex]= 3x2-2xy+ 2, you must have [itex]\Phi[/itex]= x3- x2y+ 2x+ h(y). The derivative of that, with respect to y is -x2+ h'(y)
    and, of course, that must equal N which is 6y2- x2+ 3, not 6y2- x3+ 3!!!

    You should have seen that you copied that incorrectly: you have
    -x2+ h'(y)= 6y2- x3+ 3 and then h'(y)= 6y2- x3+ x2+ 3 which is impossible: h is a function of y only!

    Writing it correctly, -x2+ h'(y)= 6y2- x2+ 3, h'(y)= 6y2+ 3. The x2 terms cancel which has to happen- that's the whole point of the differential being exact!

    With h'(y)= 6y2+ 3, h(y)= 2y3+ 3y and the general solution to the differential equation is
    x3- x2y+ 2x+ 2y3+ 3y= Constant.
     
    Last edited: Feb 27, 2006
  6. Feb 27, 2006 #5

    quasar987

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    :eek: I didn't noticed h(y) had x terms in it!

    mr_coffee, like HallsofIvy said, h(y) has to be a function of y. Why? Because if in fact there exists a function [itex]\psi(x,y)[/itex] such that [itex]\partial \psi / \partial x = M[/itex] and such that [itex]\psi(x,y) = \int Mdx + h[/itex], it must be that h is a function of y only because it it were a function of x too, differentiation of that last equation would give [itex]\partial \psi / \partial x = M + \partial h(x,y) / \partial x \neq M[/itex] which contradicts the first equation.

    So verifying that h is a function of y only is always a good way to check if you've made some mistakes along the way.
     
  7. Feb 27, 2006 #6
    Ahh thanks guys!
    Here i redid it, i think its right now:
    [​IMG]
     
  8. Feb 28, 2006 #7

    HallsofIvy

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    Yes, that is correct!
     
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