The differential equation of the family of curves $y = Ae^{3x} + Be^{-3x}$, where A and B are arbitrary constants, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{d^2y}{dx^2}=9y$ **
Given family: $y= A e^{3x}+B e^{-3x}$
First derivative: $y'=3A e^{3x}-3B e^{-3x}$
Second derivative: $y''=9A e^{3x}+9B e^{-3x}$
Since $9A e^{3x}+9B e^{-3x}=9y$, the differential equation is
$y''-9y=0$
The differential equation of the family is $y''-9y=0$.