Final answer:
The equation y'-y-6y=0 is likely a typo and should be corrected before proceeding. A general solution for a correct first-order linear differential equation would be of the form y=Ce^(kt). Quadratic equations and logistic curves represent different types of mathematical solutions.
Step-by-step explanation:
The differential equation provided, y'-y-6y=0, appears to be incorrectly written since the linear and the constant coefficients seem to be mixed. However, assuming the correct equation is y'-7y=0, we can find a general solution using separation of variables or an integrating factor. The solution would generally be of the form y=Ce^(kt), where k is the constant from the differential equation and C is an arbitrary constant. It would not include two additive exponential terms with different exponents as given in the question.
For an equation in the form ax² + bx + c = 0, the quadratic formula can be used to solve for the roots. This is not directly related to solving a differential equation but is a method used for finding solutions to quadratic equations.
The logistic curve mentioned could be a solution to a modified differential equation, such as one modeling population growth with a carrying capacity. The form of the logistic curve would be y = L/(1+Ce^(-kt)), where L is the carrying capacity, C is a constant determined by initial conditions, and k is a growth rate constant.