The differential equation of `y = A e^(5x) + B e^(-5x)` is

Options

A. `(d^(2)y)/(dx^(2)) = 25y`
B. `(d^(2)y)/(dx^(2)) = -25 y`
C. `(d^(2)y)/(dx^(2)) = 5y`
D. `y(d^(2)y)/(dx^(2)) = -5y`

Answer

To find the differential equation of the function \( y = A e^{5x} + B e^{-5x} \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = A e^{5x} + B e^{-5x} \]

We differentiate \( y \) to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A e^{5x}) + \frac{d}{dx}(B e^{-5x}) \]

Using the chain rule:
\[ \frac{dy}{dx} = A \cdot 5 e^{5x} + B \cdot (-5) e^{-5x} \]
\[ \frac{dy}{dx} = 5A e^{5x} - 5B e^{-5x} \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(5A e^{5x}) + \frac{d}{dx}(-5B e^{-5x}) \]

Again using the chain rule:
\[ \frac{d^2y}{dx^2} = 5A \cdot 5 e^{5x} + (-5B)(-5)e^{-5x} \]
\[ \frac{d^2y}{dx^2} = 25A e^{5x} + 25B e^{-5x} \]

### Step 3: Factor out common terms

We can factor out \( 25 \) from the expression:
\[ \frac{d^2y}{dx^2} = 25(A e^{5x} + B e^{-5x}) \]

Since \( A e^{5x} + B e^{-5x} = y \), we can substitute back:
\[ \frac{d^2y}{dx^2} = 25y \]

### Final Differential Equation

Thus, the differential equation is:
\[ \frac{d^2y}{dx^2} - 25y = 0 \]

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