Question

Check the condition of integrability & solve

yzdx + 2xzdy − 3xydz = 0.

Answer 1

Let us denote $U=yz\cdot i + 2xz\cdot j − 3xy\cdot k.$ Then

$curl\ U=\left|\begin{array}{ccc} i & j & k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ yz & 2xz & − 3xy \end{array}\right|=(-3x-2x)i-(-3y-y)j+(2z-z)k=$

$=-5x\cdot i +4y\cdot j+z\cdot k$

Since the scalar product $U\cdot curl U=(yz\cdot i + 2xz\cdot j − 3xy\cdot k)\cdot(-5x\cdot i +4y\cdot j+z\cdot k)=$

$-5xyz+8xyz-3xyz=0$, we conclude that $yzdx + 2xzdy − 3xydz = 0$ is integrable.

Let us divide both part of the equation $yzdx + 2xzdy − 3xydz = 0$ by $xyz$:

$\frac{dx}{x}+2\frac{dy}{y}-3\frac{dz}{z}=0$

It folows that $\int\frac{dx}{x}+2\int\frac{dy}{y}-3\int\frac{dz}{z}=0$ and therefore, $\ln |x|+2\ln|y|-3\ln|z|=\ln|C|$.

Then $\ln|xy^2|=\ln|Cz^3|$. Consequently, the solution is the following:

$xy^2=Cz^3$