The differential equation of `y=ae^(bx)` is

Options

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

Answer

To find the differential equation of the function \( y = ae^{bx} \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = ae^{bx} \]
To find the first derivative of \( y \) with respect to \( x \), we apply the chain rule:
\[ \frac{dy}{dx} = a \cdot \frac{d}{dx}(e^{bx}) = a \cdot e^{bx} \cdot b = abe^{bx} \]

### Step 2: Express \( e^{bx} \) in terms of \( y \)
From the original equation \( y = ae^{bx} \), we can express \( e^{bx} \) as:
\[ e^{bx} = \frac{y}{a} \]

### Step 3: Substitute \( e^{bx} \) into the first derivative
Now we substitute \( e^{bx} \) in the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = ab \cdot e^{bx} = ab \cdot \frac{y}{a} = by \]

### Step 4: Form the differential equation
The relationship we found is:
\[ \frac{dy}{dx} = by \]
This is the first-order linear differential equation that corresponds to the function \( y = ae^{bx} \).

### Final Answer
The differential equation of \( y = ae^{bx} \) is:
\[ \frac{dy}{dx} - by = 0 \]

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