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Show that the homogeneous equation (Ax^2+By^2)dx+(Cxy+Dy^2)d

  1. Jun 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the homogeneous equation: $$(Ax^2+By^2)dx+(Cxy+Dy^2)dy=0$$ is exact iff 2b=c.

    2. Relevant equations
    None, just definitions.

    3. The attempt at a solution

    Let $$M = Ax^2+By^2$$ and $$N = Cxy+Dy^2$$

    Taking the partial derivative of M with respect to y and the partial of N with respect to x we get

    $$\frac{\partial M}{\partial y} \ =2By$$ and $$\frac{\partial N}{\partial x} \ =Cy$$

    $$\frac{\partial M}{\partial y} \ =\frac{\partial N}{\partial x} $$ is true only if 2B=C.

    What is giving problems here is the iff statement. Can I compete this problem by stating that 2b=c, then
    $$\frac{\partial M}{\partial y} \ =\frac{\partial N}{\partial x} $$ ??

    Can someone point me in the right direction.
    Last edited by a moderator: Jun 10, 2016
  2. jcsd
  3. Jun 10, 2016 #2


    Staff: Mentor

    You have shown that if the equation is exact, then 2B = C.
    It's easy to show that if 2B = C, then the equation is exact (i.e., that ##M_y = N_x##).
    For this one, however, since each of the steps is really if and only if (reversible), you don't need to prove both statements.
  4. Jun 11, 2016 #3
    Thanks for the clarification. So it is basic substitution of 2B=C in the equation to get ##M_y = N_x##). Thanks, Mark. I was overthinking the question.
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