Answer to Question #206473 in Differential Equations for Keem

Question

Verify that the differential equation (y^2+yz)dx+(xz+z^2)dy+(y^2-xy)dz=0 is integrable and find its primitive

Answer 1

The necessary and sufficient condition for iintegrability is

$$X\cdot curl X=0$$

$X=(y^2+yz,xz+z^2,y^2-xy)$ so that

$$\nabla\times X=\begin{vmatrix} \vec i & \vec j & \vec k \\ {\partial\over \partial x} & {\partial\over \partial y} & {\partial\over \partial z} \\ y^2+yz & xz+z^2 & y^2-xy \end{vmatrix}$$

$$=(2y-x-x-2z)\vec i+(y+y)\vec j + (z-2y-z)\vec k$$

$$=(2y-2x-2z)\vec i+(2y)\vec j + (-2y)\vec k$$

$$X\cdot (\nabla\times X)=2y^3-2xy^2-2y^2z+2y^2z$$

$$-2xyz-2yz^2+2xyz+2yz^2-2y^3+2xy^2=0$$

Thus the given equation is integrable.

Solve by Inspection

$$y(y+z)dx+z(x+z)dy+y(y-x)dz=0$$

Or

$$y(y+z)dx+y(y+z)dz-y(y+z)dz$$

$$+z(x+z)dy+y(x+z)dy-y(x+z)dy$$

$$+y(y-x)dz=0$$

Or

$$y(y+z)d(x+z)+(y+z)(x+z)dy$$

$$-ydz(y+z-y+x)-y(x+z)dy=0$$

Or

$$y(y+z)d(x+z)+(y+z)(x+z)dy-y(x+z)d(y+z)=0$$

$${d(x+z) \over x+z}+{dy\over y}-{d(y+z) \over y+z}=0$$

The complete primitive is given as

$$y(x+z)=c(y+z)$$