Differential Equations Question 132
Question: $ y=ae^{mx}+b{e^{-mx}} $ satisfies which of the following differential equations
[Karnataka CET 2002]
Options:
A) $ \frac{dy}{dx}-my=0 $
B) $ \frac{dy}{dx}+my=0 $
C) $ \frac{d^{2}y}{dx^{2}}+m^{2}y=0 $
D) $ \frac{d^{2}y}{dx^{2}}-m^{2}y=0 $
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Answer:
Correct Answer: D
Solution:
$ y=ae^{mx}+b{e^{-mx}} $ . Differentiating, we get $ \frac{dy}{dx}=mae^{mx}-mb{e^{-mx}} $ . Differentiating again, we get $ \frac{d^{2}y}{dx^{2}}=m^{2}ae^{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
