Question

The equation \(\dfrac{d^2y}{dx^2} - y = 0\) has a solution of the form \(y = e^{Ax}\). The value(s) of \(A\) satisfying this is/are:

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(-\infty\)

Hint:

Substitute \(y=e^{Ax}\) into the equation to get the characteristic equation \(A^2-1=0\), then solve for A.

Answer 1

The Correct Option is B, C

Step 1: Substitute the trial solution into the equation.
We are told \(y = e^{Ax}\) is a solution of \(\dfrac{d^2y}{dx^2} - y = 0\). Differentiate once: \(\dfrac{dy}{dx} = A e^{Ax}\). Differentiate again: \(\dfrac{d^2y}{dx^2} = A^2 e^{Ax}\).

Step 2: Substitute back into the differential equation.
\[ A^2 e^{Ax} - e^{Ax} = 0 \] Factor out \(e^{Ax}\), which is never zero for any finite \(A\) and \(x\): \[ e^{Ax}(A^2 - 1) = 0 \]
Step 3: Solve the resulting algebraic equation.
Since \(e^{Ax} \neq 0\), we need \[ A^2 - 1 = 0 \] \[ A^2 = 1 \] \[ A = \pm 1 \] This is the characteristic equation of the differential equation, and its two roots give the two independent exponential solutions.

Step 4: Check each option against \(A = \pm 1\).
\(A = 0\) gives \(y = e^{0} = 1\), a constant. Substituting back: \(\dfrac{d^2}{dx^2}(1) - 1 = 0 - 1 = -1 \neq 0\), so \(A=0\) fails. \(A = 1\) gives \(A^2 - 1 = 1 - 1 = 0\), which satisfies the equation. \(A = -1\) gives \(A^2 - 1 = 1 - 1 = 0\), which also satisfies the equation. \(A \to -\infty\) is not a finite value and does not satisfy \(A^2=1\), so it fails.

Final Answer:
Both \(A=1\) and \(A=-1\) satisfy the equation. \[ \boxed{A = 1 \text{ and } A = -1} \]