Question

Elimination of arbitrary constants $A$ and $B$ from $y = Ae^x + Be^{-2x}$ gives the differential equation:

• A. $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$
• B. $\frac{d^2y}{dx^2} + \frac{dy}{dx} + 2y = 0$
• C. $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0$
• D. $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$
• E. $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0$

Hint:

Exponential solutions directly give roots of characteristic equation.

Answer 1

The Correct Option is A

Concept:
• Use characteristic equation method

Step 1: Given solution form
\[ y = Ae^x + Be^{-2x} \] Roots: $m=1$ and $m=-2$

Step 2: Form characteristic equation
\[ (m-1)(m+2) = 0 \] \[ m^2 + m - 2 = 0 \]

Step 3: Convert to differential equation
\[ \frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0 \] Final Conclusion:
Option (A)