The given equation is f(x+y) + f(x-y) = 2f(x)f(y) and the initial condition is f(0) = k.
The general solution for f(x) is f(x) = c*cos(x) + d*sin(x), where c and d are constants.
Substituting x = 0 and f(0) = k into the general solution, we get k = c*cos(0) + d*sin(0) = c. This means that c = k. Substituting this into the general solution, we get f(x) = k*cos(x) + d*sin(x). To find d, we can differentiate both sides of the equation with respect to x and substitute x = 0 to get d = f'(0).
Substituting c = k and d = f'(0) into the general solution, we get f(x) = k*cos(x) + f'(0)*sin(x). This is the specific solution for f(x) with the given initial condition.
Yes, this equation and initial condition can have infinitely many solutions because there are infinitely many possible values for c and d. Each pair of values for c and d will result in a different solution for f(x).