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The problem in the image involves finding the value of \( c \) for which the solution of the differential equation \( x(t) = Ae^{ct} - Be^{-2t} \) satisfies the given conditions.

Let's analyze the given conditions step-by-step:

### Given:
1. \( x(0) = A - B = 2 \)
2. The derivative \( x'(t) = cAe^{ct} + 2Be^{-2t} \)
3. \( x'(0) = cA + 2B = 12 \)

### Solution Steps:

1. **Solve for \( A \) and \( B \):**

   From the equation \( x(0) = A - B = 2 \), we have:
   \[
   A - B = 2
   \]

2. **Set up the system of equations:**

   From the derivative condition \( x'(0) = cA + 2B = 12 \), we have:
   \[
   cA + 2B = 12
   \]

3. **Substitute \( A \) from the first equation into the second:**

   \( A = B + 2 \). Substitute this into the second equation:

   \[
   c(B + 2) + 2B = 12
   \]

   Expanding this equation:

   \[
   cB + 2c + 2B = 12
   \]

   Combine like terms:

   \[
   (c + 2)B + 2c = 12
   \]

4. **Solve for \( B \):**

   Rearrange to solve for \( B \):

   \[
   B = \frac{12 - 2c}{c + 2}
   \]

5. **Plug \( B \) back into the expression for \( A \):**

   \[
   A = B + 2 = \frac{12 - 2c}{c + 2} + 2
   \]

6. **Analyze the choices of \( c \):**

   The options for \( c \) are \( \frac{1}{2}, -4, 0.2, \) and \( -0.4 \). We need to substitute these values into our equations to check which one satisfies both equations.

I will now compute this systematically.For all the given choices of \( c \) (\( \frac{1}{2}, -4, 0.2, -0.4 \)), both equations \( A - B = 2 \) and \( cA + 2B = 12 \) are satisfied. Thus, each value of \( c \) is a valid solution for the given conditions.

### Conclusion:
All the options provided (\( \frac{1}{2}, -4, 0.2, -0.4 \)) are correct answers.

Would you like more details on the calculations or explanations? Here are some related questions for further exploration:

1. What would be the effect if the initial conditions were different?
2. How would the solution change if the differential equation had different coefficients?
3. Can you derive the general solution for a second-order differential equation?
4. How does the characteristic equation help in solving differential equations?
5. What are some applications of solving differential equations in real-world problems?

**Tip:** When solving differential equations, always check if all provided conditions are satisfied by substituting the found solutions back into the original conditions.

Solve for c: x(t) = Ae^ct - Be^-2t, given that x(0) = A - B = 2, and the derivative x'(t) = cAe^ct + 2Be^-2t, with x'(0) = cA + 2B = 12.

This problem involves solving a differential equation of the form x(t) = Ae^ct - Be^-2t, subject to initial conditions x(0) = A - B = 2 and x'(0) = cA + 2B = 12. The steps involve algebraic manipulation and solving for constants.

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