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Differential Equations: exact equation

  1. Sep 12, 2007 #1
    i fell asleep when the professor went over how to solve exact equations :-/ i know it's really easy but despite reading the chapter over and over i still can't get it right. please show me where i'm going wrong / what to do next.

    1. The problem statement, all variables and given/known data
    Determine whether the equation in problem 1 is exact. If it is exact, find the solution.
    [tex](2x + 3) + (2y - 2)y' = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex](2x + 3)dx + (2y - 2)dy = 0[/tex]

    [tex]M_{y} = 0 = N_{x} = 0[/tex] <--- the equation is exact

    [tex]\psi_{x} = 0[/tex] --> [tex]\psi = \int^x 0dx = x + h(y)[/tex]

    [tex]\frac{d\psi}{dy} = h'(y) = 2y - 2[/tex] ---> [tex]h(y)= y^2 - 2y[/tex]

    and then i get stuck. i'm not sure where to go from there. the answer to the problem is [tex]x^2 + 3x + y^2 - 2y = c[/tex], which is apparent to me if you turn the original equation into a separable one, but that's not possible with all exact equations.

    thanks for your time everyone.
  2. jcsd
  3. Sep 12, 2007 #2


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    An exact equation is one of the form dF(x,y)=0. So it has solutions F(x,y)=c. The dF equation
    becomes F_x*dx+F_y*dy=0, where the underscores are partial derivatives. So F_x is your M and F_y is your N. What you want to do is find F. Since F_x=2x+3, F=x^2+3x+h(y). And sure F_y is then h'(y) which you know is 2y-2. Solve for h(y) (as you did). Now what is F? Hence what are the solutions?
  4. Jul 16, 2009 #3
    You already solved it. Just plug your c(y) into F. Your F is x^2 + 3x + c(y) after you integrated the M term. Plug c(y) back into F and you get x^2+3x+y^2-2y.
  5. Jul 16, 2009 #4


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    It's not really worth while recycling threads from 2007. Now is it?
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