The function `f(x) ` satisfying the equation `f^2 (x) + 4 f'(x) f(x) + (f'(x))^2 = 0`

Options

A. `f(x)=C.e^((2-sqrt3)x`
B. `f(x)=C.e^((2+sqrt3)x`
C. `f(x)=C.e^((sqrt3-2)x`
D. `f(x)=C.e^(-(2+sqrt3)x`

Answer

To solve the differential equation \( f^2(x) + 4 f'(x) f(x) + (f'(x))^2 = 0 \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ f^2(x) + 4 f'(x) f(x) + (f'(x))^2 = 0 \]
This can be rearranged as:
\[ (f'(x))^2 + 4 f(x) f'(x) + f^2(x) = 0 \]

### Step 2: Substitute \( f'(x) \)
Let \( f'(x) = t \). Then, we can rewrite the equation as:
\[ t^2 + 4 f(x) t + f^2(x) = 0 \]
This is a quadratic equation in \( t \).

### Step 3: Use the Quadratic Formula
For the quadratic equation \( at^2 + bt + c = 0 \), the solutions for \( t \) are given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 4f(x) \), and \( c = f^2(x) \). Plugging these values into the formula gives:
\[ t = \frac{-4f(x) \pm \sqrt{(4f(x))^2 - 4 \cdot 1 \cdot f^2(x)}}{2 \cdot 1} \]
This simplifies to:
\[ t = \frac{-4f(x) \pm \sqrt{16f^2(x) - 4f^2(x)}}{2} \]
\[ t = \frac{-4f(x) \pm \sqrt{12f^2(x)}}{2} \]
\[ t = \frac{-4f(x) \pm 2\sqrt{3}f(x)}{2} \]
\[ t = -2f(x) \pm \sqrt{3}f(x) \]

### Step 4: Express \( f'(x) \)
Thus, we have:
\[ f'(x) = -2f(x) + \sqrt{3}f(x) \quad \text{or} \quad f'(x) = -2f(x) - \sqrt{3}f(x) \]
This simplifies to:
\[ f'(x) = (-2 + \sqrt{3})f(x) \quad \text{or} \quad f'(x) = (-2 - \sqrt{3})f(x) \]

### Step 5: Separate Variables
For the first case:
\[ \frac{f'(x)}{f(x)} = -2 + \sqrt{3} \]
Integrating both sides gives:
\[ \ln |f(x)| = (-2 + \sqrt{3})x + C \]
Exponentiating both sides results in:
\[ f(x) = e^{C} e^{(-2 + \sqrt{3})x} \]
Let \( C_1 = e^{C} \), then:
\[ f(x) = C_1 e^{(-2 + \sqrt{3})x} \]

For the second case:
\[ \frac{f'(x)}{f(x)} = -2 - \sqrt{3} \]
Integrating gives:
\[ \ln |f(x)| = (-2 - \sqrt{3})x + C \]
Exponentiating results in:
\[ f(x) = C_2 e^{(-2 - \sqrt{3})x} \]

### Final Solution
Thus, the general solution for the function \( f(x) \) is:
\[ f(x) = C e^{(-2 + \sqrt{3})x} \quad \text{or} \quad f(x) = C e^{(-2 - \sqrt{3})x} \]
where \( C \) is a constant.