Consider the ODE y′+3y=0y' + 3y = 0y′+3y=0. Which function y(x)y(x)y(x) satisfies y(0)=2y(0) = 2y(0)=2?
y=2e3xy = 2e^{3x}y=2e3x
y=2e−3xy = 2e^{-3x}y=2e−3x
y=e−3x+1y = e^{-3x} + 1y=e−3x+1
y=3e−2xy = 3e^{-2x}y=3e−2x