Differential Equations Question 336
Question: The differential equation of the family of curves $ y=Ae^{3x}+Be^{5x}, $ where A and B are arbitrary constants, is
[MNR 1988]
Options:
A) $ \frac{d^{2}y}{dx^{2}}+8\frac{dy}{dx}+15y=0 $
B) $ \frac{d^{2}y}{dx^{2}}-8\frac{dy}{dx}+15y=0 $
C) $ \frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}+y=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=Ae^{3x}+Be^{5x} $
Therefore $ \frac{dy}{dx}=3Ae^{3x}+5Be^{5x} $
Therefore $ \frac{d^{2}y}{dx^{2}}=9Ae^{3x}+25Be^{5x} $
Therefore $ \frac{d^{2}y}{dx^{2}}-8\frac{dy}{dx}+15y=0 $ (By inspection)
BETA
