Should be easy for Mathematicians. The partial deferential equation is from Thermodynamics and the book presents the following without a worked sequence. I have taken Calculus 3 and differential equations (introduction to ordinary differential equations).
$G=H+(\partial G/\partial T)T$
"can be rewritten as"
$\partial /\partial T(G/T)=-H/T^2$
Can someone show me the work? I have tried to separate variables and integrate but I get
$\dfrac{1}{G-H}\partial G=\dfrac{\partial T}{T}$
Integrating the left gives a logarithm so I know this can't be the correct procedure.
Thanks, Mathematicians.
By the quotient rule,
$$\frac{\partial}{\partial T}(G/T) = \frac{T \frac{\partial G}{\partial T} - G \frac{\partial T}{\partial T}}{T^2} = \frac{\partial G \over \partial T}{T} - \frac{G}{T^2}$$
Hence $\displaystyle \frac{\partial}{\partial T}(G/T)=-H/T^2$ is equivalent to
$$T \frac{\partial G}{\partial T} - G = -H$$
which is equivalent to your original expression.