If the solution of the equation `(d^(2)x)/(dt^(2))+4(dx)/(dt)+3x = 0` given that for `t = 0, x = 0 and (dx)/(dt) = 12` is in the form `x = Ae^(-3t) + Be^(-t)`, then

Options

A. `A + B = 0`
B. `A + B = 12`
C. `|AB| = 36`
D. `|AB| = 49`

Explanation

To solve the differential equation

\[ \frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 3x = 0 \]

given the initial conditions \( x(0) = 0 \) and \( \frac{dx}{dt}(0) = 12 \), we start with the proposed solution form:

\[ x = Ae^{-3t} + Be^{-t} \]

### Step 1: Apply the initial condition \( x(0) = 0 \)

Substituting \( t = 0 \) into the solution:

\[ x(0) = A e^{0} + B e^{0} = A + B \]

Since \( x(0) = 0 \):

\[ A + B = 0 \quad \text{(Equation 1)} \]

### Step 2: Differentiate \( x \) to find \( \frac{dx}{dt} \)

Differentiating \( x \):

\[ \frac{dx}{dt} = -3Ae^{-3t} - Be^{-t} \]

### Step 3: Apply the initial condition \( \frac{dx}{dt}(0) = 12 \)

Substituting \( t = 0 \) into the derivative:

\[ \frac{dx}{dt}(0) = -3A e^{0} - B e^{0} = -3A - B \]

Since \( \frac{dx}{dt}(0) = 12 \):

\[ -3A - B = 12 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations

From Equation 1, we have:

\[ B = -A \]

Substituting \( B = -A \) into Equation 2:

\[ -3A - (-A) = 12 \]

This simplifies to:

\[ -3A + A = 12 \]
\[ -2A = 12 \]
\[ A = -6 \]

Now substituting \( A = -6 \) back into Equation 1 to find \( B \):

\[ -6 + B = 0 \implies B = 6 \]

### Step 5: Conclusion

We have found:

\[ A = -6, \quad B = 6 \]

### Step 6: Verify the conditions

1. **Check \( A + B = 0 \)**:
\[ -6 + 6 = 0 \quad \text{(True)} \]

2. **Check \( |A \cdot B| = 36 \)**:
\[ |-6 \cdot 6| = 36 \quad \text{(True)} \]

3. **Check \( A + B = 12 \)**:
\[ -6 + 6 = 0 \quad \text{(False)} \]

Thus, the correct options are:

- \( A + B = 0 \)
- \( |A \cdot B| = 36 \)