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

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

with the initial conditions \( x(0) = 0 \) and \( \frac{dx}{dt}(0) = 12 \), we will follow these steps:

### Step 1: Find the characteristic equation
The given differential equation is a second-order linear homogeneous equation. We can find the characteristic equation by assuming a solution of the form \( x = e^{rt} \). Substituting this into the differential equation gives:

\[ r^2 + 4r + 3 = 0 \]

### Step 2: Solve the characteristic equation
Next, we will solve the quadratic equation \( r^2 + 4r + 3 = 0 \) using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1, b = 4, c = 3 \).

Calculating the discriminant:

\[ b^2 - 4ac = 4^2 - 4 \times 1 \times 3 = 16 - 12 = 4 \]

Now substituting into the quadratic formula:

\[ r = \frac{-4 \pm \sqrt{4}}{2 \times 1} = \frac{-4 \pm 2}{2} \]

This gives us the roots:

\[ r_1 = \frac{-2}{2} = -2 \quad \text{and} \quad r_2 = \frac{-6}{2} = -3 \]

### Step 3: Write the general solution
The general solution of the differential equation is given by:

\[ x(t) = A e^{-2t} + B e^{-3t} \]

### Step 4: Apply initial conditions
We are given the initial conditions \( x(0) = 0 \) and \( \frac{dx}{dt}(0) = 12 \).

1. **Using \( x(0) = 0 \)**:
\[ x(0) = A e^{0} + B e^{0} = A + B = 0 \quad \text{(Equation 1)} \]

2. **Finding \( \frac{dx}{dt} \)**:
Differentiate \( x(t) \):
\[ \frac{dx}{dt} = -2A e^{-2t} - 3B e^{-3t} \]

Now apply the second initial condition \( \frac{dx}{dt}(0) = 12 \):
\[ \frac{dx}{dt}(0) = -2A - 3B = 12 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
We have the following system of equations:

1. \( A + B = 0 \)
2. \( -2A - 3B = 12 \)

From Equation 1, we can express \( B \) in terms of \( A \):
\[ B = -A \]

Substituting into Equation 2:
\[ -2A - 3(-A) = 12 \implies -2A + 3A = 12 \implies A = 12 \]

Now substituting \( A = 12 \) back into Equation 1:
\[ 12 + B = 0 \implies B = -12 \]

### Step 6: Write the final solution
Thus, the solution to the differential equation is:

\[ x(t) = 12 e^{-2t} - 12 e^{-3t} \]