If \(y = e^{kx}\) is a solution of the differential equation \(\displaystyle\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y = 0\), then the possible values of k are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3 and −2
A function is a solution when it makes the equation true for every x, so the derivatives are put into the equation.
\( \displaystyle \frac{dy}{dx} = ke^{kx}, \quad \frac{d^2y}{dx^2} = k^2e^{kx} \)
Substituting:
\(k^2e^{kx} - ke^{kx} - 6e^{kx} = 0\)
\(e^{kx}(k^2 - k - 6) = 0\)
Since \(e^{kx}\) is never zero:
\(k^2 - k - 6 = 0\)
\((k - 3)(k + 2) = 0\)
\(k = 3 \text{ or } k = -2\)