Question

Find the differential equation by eliminating arbitrary constants from the relation \( y = Ae^{3x} + Be^{-3x} \).

Hint:

The order of the resulting differential equation is equal to the number of independent arbitrary constants in the original relation. Differentiate that many times and then use algebra to eliminate the constants.

Answer 1

Since there are two arbitrary constants (\(A\) and \(B\)), we need to differentiate the relation twice.
Step 1: Differentiate with respect to \(x\). \[ \frac{dy}{dx} = 3Ae^{3x} - 3Be^{-3x} \quad \cdots (1) \] Step 2: Differentiate again with respect to \(x\). \[ \frac{d^2y}{dx^2} = 9Ae^{3x} + 9Be^{-3x} \quad \cdots (2) \] Step 3: Eliminate the constants.
Factor out 9 from the right side of equation (2): \[ \frac{d^2y}{dx^2} = 9(Ae^{3x} + Be^{-3x}) \] Notice that the expression in the parentheses is the original expression for \(y\). \[ \frac{d^2y}{dx^2} = 9y \] The required differential equation is \( \frac{d^2y}{dx^2} - 9y = 0 \).