If differential equation of the curve $y = ae^{3x} + be^{2x} + ce^x$ is $\frac{d^3y}{dx^3} + m\frac{d^2y}{dx^2} + n\frac{dy}{dx} + p y = 0$, then:
$(m + n)$ is a prime number
$(m - n)$ is always divisible by any odd number
$m + n + p$ is a negative integer
$|m|$ is divisible by two prime numbers
Step-by-Step Solution
Key Concept: For the family of curves y = ae^(3x) + be^(2x) + ce^x with three arbitrary constants, the differential equation must be third-order. The roots of the characteristic equation r³ - 6r² + 11r - 6 = 0 are 1, 2, 3 (matching the exponents), yielding coefficients m = -6, n = 11, p = -6.
The given differential equation is $\frac{d^3y}{dx^3} - 6\frac{d^2y}{dx^2} + 11\frac{dy}{dx} - 6y = 0$. This is a linear homogeneous differential equation with constant coefficients. The characteristic equation is obtained by substituting $y = e^{mx}$, giving $m^3 - 6m^2 + 11m - 6 = 0$, which factors to $(m-1)(m-2)(m-3) = 0$, yielding roots $m = 1, 2, 3$. The general solution is $y = c_1e^x + c_2e^{2x} + c_3e^{3x}$.
Correct Answer: 1,3,4