Question

Verify that the total differential equation z(z-y)dx+z(x+z)dy+x(x+y)dz=0 is integrable and hence find its integral

Answer 1

The integrability condition:

$𝑃 ( ∂𝑄/ ∂𝑧 − ∂𝑅/ ∂𝑦 ) + 𝑄 ( ∂𝑅/ ∂𝑥 − ∂𝑃/ ∂𝑧 ) + 𝑅 ( ∂𝑃/ ∂𝑦 − ∂𝑄/ ∂𝑥 ) = 0$

$z(z-y)(x+2z-x)+z(x+z)(2x+y-2z+y)+x(x+y)(-z-z)=$

$=2z^2(z-y)+2z(x+z)(x+y-z)-2zx(x+y)=$

$=2z^3-2z^2y+2zx^2+2zxy-2z^2x+2z^2x+2z^2y-2z^3-2zx^2-2xyz=0$

The equation is integrable.

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

This equation is homogeneous.

$Px+Qy+Rz=z(z-y)x+z(x+z)y+x(x+y)z=$

$=z^2x+z^2y+x^2z+xyz=D\neq0$

Integrating factor:

$\frac{1}{D}=\frac{1}{z^2x+z^2y+x^2z+xyz}$

Then:

$\frac{z(z-y)dx+z(x+z)dy+x(x+y)dz}{z^2x+z^2y+x^2z+xyz}=\frac{z(z-y)dx+z(x+z)dy+x(x+y)dz}{z(x+y)(x+z)}=0$

$\frac{(z-y)dx}{(x+y)(x+z)}+\frac{dy}{x+y}+\frac{xdz}{z(x+z)}=0$

$d(ln(x+y)-ln(x+z))+\frac{dz}{x+z}+\frac{xdz}{z(x+z)}=0$

$d(ln(x+y)-ln(x+z))+\frac{(x+z)dz}{z(x+z)}=0$

$ln(x+y)-ln(x+z)+lnz=lnC$

$\frac{z(x+y)}{x+z}=C$