$\left(3x^2y-xy\right)dx+\left(2x^3y^2+x^3y^4\right)dy=0$
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?
Best Answer
$$\left(3x^2y-xy\right)dx+\left(2x^3y^2+x^3y^4\right)dy=0$$ $$xy\Big(\left(3x-1\right)dx+x^2y\left(2+y^2\right)dy\Big)=0$$ 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.