The differential equation for `y=ae^(x)+be^(-2x)` is

Options

A. `(d^(2)y)/(dx^(2))+(dy)/(dx)-2y=0`
B. `(d^(2)y)/(dx^(2))-(dy)/(dx)-2y=0`
C. `(d^(2)y)/(dx^(2))+(dy)/(dx)-2y=0`
D. `(d^(2)y)/(dx^(2))-(dy)/(dx)+2y=0`

Answer

To find the differential equation for the function \( y = ae^x + be^{-2x} \), we will follow these steps:

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

Given:
\[ y = ae^x + be^{-2x} \]

We differentiate \( y \) to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{d}{dx}(ae^x) + \frac{d}{dx}(be^{-2x}) \]
Using the rules of differentiation:
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( e^{-2x} \) is \( -2e^{-2x} \).

Thus:
\[ \frac{dy}{dx} = ae^x - 2be^{-2x} \]

### Step 2: Express \( e^{-2x} \) in terms of \( y \) and \( \frac{dy}{dx} \)

From the expression for \( y \):
\[ be^{-2x} = y - ae^x \]

### Step 3: Substitute \( be^{-2x} \) into the derivative

Now substitute \( be^{-2x} \) into the derivative:
\[ \frac{dy}{dx} = ae^x - 2(y - ae^x) \]
This simplifies to:
\[ \frac{dy}{dx} = ae^x - 2y + 2ae^x \]
Combining like terms:
\[ \frac{dy}{dx} = 3ae^x - 2y \]

### Step 4: Differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3ae^x - 2y) \]
Using the product rule and chain rule:
\[ \frac{d^2y}{dx^2} = 3ae^x - 2\frac{dy}{dx} \]

### Step 5: Substitute \( \frac{dy}{dx} \) back into the second derivative

Now substitute \( \frac{dy}{dx} = 3ae^x - 2y \) into the second derivative:
\[ \frac{d^2y}{dx^2} = 3ae^x - 2(3ae^x - 2y) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = 3ae^x - 6ae^x + 4y \]
Combining like terms gives:
\[ \frac{d^2y}{dx^2} = -3ae^x + 4y \]

### Step 6: Rearranging to form the differential equation

Rearranging the equation gives us:
\[ \frac{d^2y}{dx^2} + 3ae^x - 4y = 0 \]

### Final Differential Equation

Thus, the final differential equation is:
\[ \frac{d^2y}{dx^2} - 4y + 3ae^x = 0 \]