x∈R, then find ab.  Suppose f(x)=eax+ebx where a≠b and f′′(x)−2f1(x)−15f(x)=0 for all

f(x)=aex+bex⇒f′(x)=aeax+bebx,f"(x)=a2eax+b2ebxa2eax+b2ebx−2aeax+bebx−15eax+ebx=0a2−2a−15eax+b2−2b−15ebx=0   ∀xa2−2a−15=0    b2−2b−15=0a=-5,    b=5,b=−3a≠bab=5(−3)=−15