PX153 - G2 - the total differential, and exact and inexact differentials

df=∂f∂xdx+∂f∂ydy df=A(x,y)dx+B(x,y)dy

- this is no longer explicitly l\inked to a function, f(x,y)

A(x,y)=∂f∂xandB(x,y)=∂f∂y$$suchthat:$$∂A∂y=∂2f∂y∂x=∂2f∂x∂y=∂B∂x ∂A∂y=∂B∂x df=∑i=1ngi(x1...xn)dxi ∂gi∂xj=∂gj∂xi$$forallpairsof$i$and$j$−eg:showthatthefollowingpairsofdifferentialisexactandfindthefunctionfromwhichthedifferentialisderived:$$(y+z).dx+x.dy+x.dzdf=gx.dx+gy.dy+gz.dz

gx=y+z, gy=x, and gz=x
∂gx∂y=1=∂gy∂x, ∂gx∂z=1=∂gz∂x, ∂gy∂x=0=∂gz∂y
- so, it is exact:
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$$
$$\frac{\partial f}{\partial x} = y+z \implies f(x,y,z) = x(y+z)+h_{x}(y,z)+c$$
$$\frac{df}{dy}= x \implies f(x,y,z) = xy + h_{y}(x,z) +c$$
$$\frac{df}{dz}= x \implies f(x,y,z) = xz + h_{z}(x,y) +c$$
- these are satisfied by: f(x,y,z)=x(y+z)+c

- the constant of integration could involve other variables held constant