Question

\(y=ae^{mx}+be^{-mx}\) is the solution of the differential equation:

• A. \(\dfrac{dy}{dx}-my=0\)
• B. \(\dfrac{dy}{dx}+my=0\)
• C. \(\dfrac{d^2y}{dx^2}+m^2y=0\)
• D. \(\dfrac{d^2y}{dx^2}-m^2y=0\)

Hint:

If the solution is a linear combination of exponentials \(e^{\pm mx}\), the characteristic equation roots are \(\pm m\). Hence the ODE is \(y''-m^2y=0\).

Answer 1

The Correct Option is D

Step 1: Differentiate once.
\[ y=ae^{mx}+be^{-mx} \] \[ \frac{dy}{dx}=am e^{mx}-bm e^{-mx}. \]

Step 2: Differentiate again.
\[ \frac{d^2y}{dx^2}=a m^2 e^{mx}+b m^2 e^{-mx}. \]

Step 3: Compare with \(y\).
Notice that: \[ \frac{d^2y}{dx^2}=m^2\big(ae^{mx}+be^{-mx}\big)=m^2y. \]

Step 4: Rearrange to standard form.
\[ \frac{d^2y}{dx^2}-m^2y=0. \]

Final Answer:
\[ \boxed{\dfrac{d^2y}{dx^2}-m^2y=0} \]