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

1. Jun 10, 2016

### MidgetDwarf

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. Jun 10, 2016

### 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.

3. Jun 11, 2016

### MidgetDwarf

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.