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: Rewrite the equation using the operator D
We can express the differential equation in terms of the differential operator \( D \), where \( D = \frac{d}{dx} \). Thus, we rewrite the equation as:

\[ D^2y - 2Dy + y = 0. \]

### Step 2: Factor the operator
Next, we can factor the left-hand side of the equation:

\[ (D^2 - 2D + 1)y = 0. \]

### Step 3: Solve the characteristic equation
To find the general solution, we need to solve the characteristic equation:

\[ D^2 - 2D + 1 = 0. \]

This can be factored as:

\[ (D - 1)(D - 1) = 0, \]

which simplifies to:

\[ (D - 1)^2 = 0. \]

### Step 4: Determine the roots
The roots of the characteristic equation are both \( D = 1 \) (a repeated root).

### Step 5: Write the general solution
For a repeated root \( r \), the general solution of the differential equation is given by:

\[ y = (C_1 + C_2 x)e^{rx}, \]

where \( C_1 \) and \( C_2 \) are constants, and \( r \) is the root. In our case, since \( r = 1 \), we have:

\[ y = (C_1 + C_2 x)e^{1 \cdot x} = (C_1 + C_2 x)e^x. \]

### Conclusion
Thus, the general solution of the differential equation

\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0 \]

is

\[ y = (C_1 + C_2 x)e^x. \]