Differential Equations Question 252
Question: If $ y=e^{4x}+2{e^{-x}} $ satisfies the relation $ \frac{d^{3}y}{dx^{3}}+A\frac{dy}{dx}+By=0, $ then values of A and B respectively are:
Options:
A) -13, 14
B) -13, -12
C) -13, 12
D) 12, -13
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given $ y=e^{4x}+2{e^{-x}} $ Differentiating we get $ \frac{dy}{dx}=4e^{4x}-2{e^{-x}}\Rightarrow \frac{d^{2}y}{dx^{2}}=16e^{14x}+2{e^{-x}} $
$ \Rightarrow \frac{d^{3}y}{dx^{3}}=64e^{4x}-2{e^{-x}} $ Putting these values in $ \frac{d^{3}y}{dx^{3}}+A\frac{dy}{dx}+By=0 $ We have, $ (64+4A+B)e^{4x}+(-2-2A+2B){e^{-x}}=0 $
$ \Rightarrow 64+4A+B=0,-2-2A+2B=0 $ Solving these eqs. we get $ A=-13,B=-12 $
BETA
