Differentiation Question 43
Question: If $ y=ae^{mx}+b{e^{-mx}} $ , then $ \frac{d^{2}y}{dx^{2}}-m^{2}y $ is equal to
Options:
A) $ m^{2}(ae^{mx}-b{e^{-mx}}) $
B) 1
C) 0
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=ae^{mx}+b{e^{-mx}} $
$ \therefore \frac{dy}{dx}=ame^{mx}=mb{e^{-mx}} $ Again $ \frac{d^{2}y}{dx^{2}}=am^{2}e^{mx}+m^{2}b{e^{-mx}} $
$ =m^{2}(ae^{mx}+b{e^{-mx}})=m^{2}y $ Or $ \frac{d^{2}y}{dx^{2}}-m^{2}y=0 $
BETA
