$\left(3x^2y-xy\right)dx+\left(2x^3y^2+x^3y^4\right)dy=0$
I'm trying to solve this first-order differential equation. I know it's not an exact equation so I'm trying to use the method taught in class to solve it. I get stuck trying to do the integrating factor. Below is what I have:
$\frac{\partial \:}{\partial \:x}\left(M\right)=\frac{\partial }{\partial x}\left(3x^2y-xy\right) = 6xy-y$
$\frac{\partial \:}{\partial y}\left(N\right)=\frac{\partial }{\partial y}\left(2x^3y^2+x^3y^4\right)dy = 4x^3y+4x^3y^3$
And since $\frac{\partial \:}{\partial \:y}\left(N\right)\neq\frac{\partial }{\partial \:x}\left(M\right)$, have that it is not exact.
So we apply the formula to get an integrating factor: $\xi =\frac{\left(\:\frac{\partial \:}{\partial \:y}-\frac{\partial }{\partial x}\right)}{N}$ to get a function $\xi(x)$, or the formula $\xi =\frac{\left(\:\frac{\partial \:}{\partial \:y}-\frac{\partial }{\partial x}\right)}{-M}\:$ to get a function $\xi(y)$.
We use $\xi$ to get an integrating factor $\mu(x)=e^{\int \:\xi(x) dx}$ or $\mu(y)=e^{\int \:\xi(x) dy}$.
Now, when I apply either one of the formulas for $\xi$, I always get a result dependent on both $x$ and $y$, so I'm unable to get the integrating factor.
Is there supposed to be a simpler way to solve this? I'm using this method because it's what was taught in class, but is there another simple way to solve this that I'm not seeing?
$$\left(3x^2y-xy\right)dx+\left(2x^3y^2+x^3y^4\right)dy=0$$ $$xy\Big(\left(3x-1\right)dx+x^2y\left(2+y^2\right)dy\Big)=0$$ First trivial solution : $$y(x)=0$$ Second trivial solution : $$x(y)=0$$ Remaining equation : $$\left(3x-1\right)dx+x^2y\left(2+y^2\right)dy=0$$ Changing $y$ into $-y$ doesn't change the equation. This suggests the change of function : $$Y(x)=y^2(x)$$ $$\left(3x-1\right)dx+\frac12 x^2\left(2+Y\right)dY=0$$ The equation is separable. Solvint it for $Y(x)$ is straightforward.