y = aeᵐˣ + be⁻ᵐˣ satisfies which of the following differential equation?

The correct option is d²y/dx² - m²y = 0.

To find the differential equation satisfied by the given function, we start with the given equation:

$$y=a{e}^{mx}+b{e}^{-mx}$$

Here, $a$ and $b$ are arbitrary constants, and $m$ is a constant parameter. Since there are two arbitrary constants, we differentiate the function twice with respect to $x$ to eliminate them.

Step 1: First Differentiation
Differentiating both sides of the equation with respect to $x$, we apply the chain rule for differentiation:

$$\frac{dy}{dx}=\frac{d}{dx}\left(a{e}^{mx}+b{e}^{-mx}\right)$$

This gives:

$$\frac{dy}{dx}=a·m{e}^{mx}+b·(-m)e^{-mx}$$

Simplifying the expression:

$$\frac{dy}{dx}=m\left(a{e}^{mx}-b{e}^{-mx}\right)$$

Step 2: Second Differentiation
Differentiating both sides again with respect to $x$:

$$\frac{d^{2}y}{d{x}^2}=m·\frac{d}{dx}\left(a{e}^{mx}-b{e}^{-mx}\right)$$

Applying the differentiation rules:

$$\frac{d^{2}y}{d{x}^2}=m\left[a·m{e}^{mx}-b·(-m)e^{-mx}\right]$$

Simplifying the signs and factoring out $m$:

$$\frac{d^{2}y}{d{x}^2}=m^2\left(a{e}^{mx}+b{e}^{-mx}\right)$$

Step 3: Substitution
We notice that the term inside the parentheses is exactly our original function $y$. Therefore, we can substitute $y$ back into the equation:

$$\frac{d^{2}y}{d{x}^2}=m^{2}y$$

Step 4: Rearranging the Terms
Subtracting $m^{2}y$ from both sides gives the standard form of the differential equation:

$$\frac{d^{2}y}{d{x}^2}-m^{2}y=0$$

Thus, the function satisfies the differential equation $\frac{d^{2}y}{d{x}^2}-m^{2}y=0$.