Final answer:
The value of Beta (β) in the given second-order differential equation y′′+2y=0 is √2.
Step-by-step explanation:
The general solution to the second-order differential equation y′′+2y=0 is in the form y(x)=c1cosβx+c2sinβx. To find the value of Beta (β) where β > 0, we need to compare the given equation to the general form and determine the value of β. In this case, β would be equal to the square root of 2. Therefore, the value of Beta is √2.