For the function f(x)=cosh(x)=ex+e−x2f(x) = \cosh(x) = \frac{e^x + e^{-x}}{2}f(x)=cosh(x)=2ex+e−x, which differential equation does it satisfy?
f′′(x)+f(x)=0f''(x) + f(x) = 0f′′(x)+f(x)=0
f′′(x)−f(x)=0f''(x) - f(x) = 0f′′(x)−f(x)=0
f′(x)=f(x)2f'(x) = f(x)^2f′(x)=f(x)2
f′′(x)=1f''(x) = 1f′′(x)=1