Differential Equations: exact equation

  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. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    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. 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. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

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