Is This Differential Equation Exact or Solvable by an Integrating Factor?

[mahler1]

Homework Statement .
Solve the differential equation: $(3x^2-y^2)dy-2xydx=0$. The attempt at a solution.
I thought this was an exact differential equation. If I call $M(x,y)=-2xy$ and $N(x,y)=3x^2-y^2$, then the ODE is an exact differential equation if and only if $\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}$. Now, when I compute these two partial derivatives, $\frac{\partial M}{\partial y}=-2x$ and $\frac{\partial N}{\partial x}=6x$ which are clearly different. Am I doing something wrong or is it just that this equation is not exact?

 

[Mark44]

Homework Statement .
Solve the differential equation: $(3x^2-y^2)dy-2xydx=0$.


The attempt at a solution.
I thought this was an exact differential equation. If I call $M(x,y)=-2xy$ and $N(x,y)=3x^2-y^2$, then the ODE is an exact differential equation if and only if $\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}$. Now, when I compute these two partial derivatives, $\frac{\partial M}{\partial y}=-2x$ and $\frac{\partial N}{\partial x}=6x$ which are clearly different. Am I doing something wrong or is it just that this equation is not exact?

Looks to me to be not exact.

 

[mahler1]

Looks to me to be not exact.

Yes, I've looked up in an ODE textbook and I've found that you can reduce this equation to an exact differential equation by something called "the integrating factor". Thanks anyway.

 

[STEMucator]

I find it helps writing the equation like so to avoid negative sign errors:

$(3x^2-y^2)dy + (-2xy)dx=0$

I can see an integrating factor in your equation that relies only on $x$. Using this integrating factor, you might be able to turn this equation into an exact equation. Then it is easily solvable.

 

[mahler1]

I find it helps writing the equation like so to avoid negative sign errors:

$(3x^2-y^2)dy + (-2xy)dx=0$

I can see an integrating factor in your equation that relies only on $x$. Using this integrating factor, you might be able to turn this equation into an exact equation. Then it is easily solvable.

Thanks, I've been able to reduce the equation to an exact equation by finding $μ(x)$.