The integral of 0 does have a constant (unless the initial conditions prove otherwise). However since the answer is in the form f(x,y) = c and f(x,y) contains g(y), when g’(y)=0, it’s integral is technically 0+c but the answer being in the form f(x,y) = c already accounts for any potential constants. So, we don’t need to worry about the constant from integrating g’(y).
Isn't the integrand of 0 a random constant? Since the derivative of any constant is 0? Why do you cut it? What am I missing here?
The integral of 0 does have a constant (unless the initial conditions prove otherwise). However since the answer is in the form f(x,y) = c and f(x,y) contains g(y), when g’(y)=0, it’s integral is technically 0+c but the answer being in the form f(x,y) = c already accounts for any potential constants. So, we don’t need to worry about the constant from integrating g’(y).
Oh ok, thanks for the insight