Which of the following is a general solution of `(d^(2)y)/(dx^(2))-2(dy)/(dx)+y=0`

Options

A. `y=(Ax+B)e^(x)`
B. `y=(Ax+B)e^(-x)`
C. `y=Ax^(x)+Be^(-x)`
D. `y=Acos x+Bsinx`

Explanation

To solve the differential equation \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0\), we will follow these steps:

### Step 1: Write the Differential Equation in Standard Form
The given differential equation is already in standard form:
\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0 \]

### Step 2: Assume a Solution of the Form \(y = e^{rx}\)
We assume a solution of the form \(y = e^{rx}\), where \(r\) is a constant.

### Step 3: Substitute into the Differential Equation
Now, we need to find the first and second derivatives of \(y\):
- First derivative: \(\frac{dy}{dx} = re^{rx}\)
- Second derivative: \(\frac{d^2y}{dx^2} = r^2e^{rx}\)

Substituting these into the differential equation gives:
\[ r^2e^{rx} - 2re^{rx} + e^{rx} = 0 \]

### Step 4: Factor Out \(e^{rx}\)
Since \(e^{rx} \neq 0\), we can factor it out:
\[ e^{rx}(r^2 - 2r + 1) = 0 \]
This simplifies to:
\[ r^2 - 2r + 1 = 0 \]

### Step 5: Solve the Characteristic Equation
Now we solve the characteristic equation:
\[ r^2 - 2r + 1 = 0 \]
This can be factored as:
\[ (r - 1)^2 = 0 \]
Thus, we have a repeated root:
\[ r = 1 \]

### Step 6: Write the General Solution
Since we have a repeated root, the general solution of the differential equation is given by:
\[ y = C_1 e^{rx} + C_2 x e^{rx} \]
Substituting \(r = 1\):
\[ y = C_1 e^{x} + C_2 x e^{x} \]
where \(C_1\) and \(C_2\) are constants.

### Final Answer
Thus, the general solution of the differential equation \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0\) is:
\[ y = C_1 e^{x} + C_2 x e^{x} \]