Are two independent functions constant?

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Boltzman Oscillation
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I was reading Griffith's introduction to QM book and he finds the time independent Schrödinger equation by assuming the wave function to be the product of two independent functions. He eventually gets to this:

ih(∂ψ/∂x)/(ψ) = -(h^2/2m)*(∂''φ/∂x^2)/φ + V
he says that "the left side is a function of t alone and the right side is a function of x alone. The only way this can possibly be true is if both sides are in fact constant." Why is this true?
 
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Boltzmann Oscillation said:
I was reading Griffith's introduction to QM book and he finds the time independent Boltzmann equation by assuming the wave function to be the product of two independent functions. He eventually gets to this:

ih(∂ψ/∂x)/(ψ) = -(h^2/2m)*(∂''φ/∂x^2)/φ + V
he says that "the left side is a function of t alone and the right side is a function of x alone. The only way this can possibly be true is if both sides are in fact constant." Why is this true?
Given ##g(t)=f(x)## we have ##g(t) - g(s) = f(x)-f(x)=0##. Thus ##g(t)=g(s)## for all ##t,s## which means, that ##g(t)## is constant, e.g. ##g(t)=g(0)##.
 
fresh_42 said:
Given ##g(t)=f(x)## we have ##g(t) - g(s) = f(x)-f(x)=0##. Thus ##g(t)=g(s)## for all ##t,s## which means, that ##g(t)## is constant, e.g. ##g(t)=g(0)##.
How were you able to subtract one side by g(s) and the other side by f(x)?
 
Boltzmann Oscillation said:
How were you able to subtract one side by g(s) and the other side by f(x)?
You can also write ##t_1## and ##t_2## or ##t## and ##t'##. I just need two different values for ##t##. They both have the same right hand side ##f(x)## per premise of the statement.