Let's solve **Problem 1** and **Problem 2** step by step, with explanations at each step. --- ## **Problem 1** **Given:** \( Y_1(t) = e^{2t} \) and \( Y_2(t) = e^{3t} \) are solutions to the differential equation \( y'' + 5y' + 6y = 0 \). **Show:** They form a FUNDAMENTAL SET of solutions. --- ### **Step 1: Check if they are solutions** Plug each \( Y_1 \) and \( Y_2 \) into the differential equation: - For \( Y_1(t) = e^{2t} \): \[ Y_1' = 2e^{2t}, \quad Y_1'' = 4e^{2t} \] \[ Y_1'' + 5Y_1' + 6Y_1 = 4e^{2t} + 5(2e^{2t}) + 6e^{2t} = 4e^{2t} + 10e^{2t} + 6e^{2t} = 20e^{2t} \] But wait, let's check the coefficients: The equation is \( y'' + 5y' + 6y = 0 \), so substitute \( y = e^{rt} \): \[ r^2 + 5r + 6 = 0 \implies (r+2)(r+3)=0 \implies r = -2, -3 \] But \( e^{2t} \) and \( e^{3t} \) are not solutions! The given functions should be \( e^{-2t} \) and \( e^{-3t} \). Let's proceed with the given functions anyway: \[ \text{For } Y_1(t) = e^{2t}: \] \[ Y_1' = 2e^{2t}, \quad Y_1'' = 4e^{2t} \] \[ Y_1'' + 5Y_1' + 6Y_1 = 4e^{2t} + 10e^{2t} + 6e^{2t} = 20e^{2t} \neq 0 \] So \( Y_1(t) = e^{2t} \) **is not a solution**. The problem appears to have a typo. The correct solutions for this equation should be \( e^{-2t} \) and \( e^{-3t} \). Let's **assume the problem meant \( Y_1(t) = e^{-2t} \), \( Y_2(t) = e^{-3t} \)**. Now, \[ Y_1' = -2e^{-2t}, \quad Y_1'' = 4e^{-2t} \] \[ Y_1'' + 5Y_1' + 6Y_1 = 4e^{-2t} + 5(-2e^{-2t}) + 6e^{-2t} = 4e^{-2t} - 10e^{-2t} + 6e^{-2t} = 0 \] Thus, **both are solutions**. **Explanation:** Both functions solve the differential equation. --- ### **Step 2: Check if they are linearly independent** Compute the Wronskian: \[ W(Y_1, Y_2) = \begin{vmatrix} e^{-2t} & e^{-3t} \\ -2e^{-2t} & -3e^{-3t} \end{vmatrix} = e^{-2t}(-3e^{-3t}) - e^{-3t}(-2e^{-2t}) = -3e^{-5t} + 2e^{-5t} = -e^{-5t} \] Since \( W(Y_1, Y_2) \neq 0 \) for all \( t \), they are linearly independent. **Explanation:** The Wronskian is nonzero, so the functions are linearly independent and form a fundamental set. --- ### **Final Answer for Problem 1:** **Yes, \( Y_1(t) = e^{-2t} \) and \( Y_2(t) = e^{-3t} \) form a fundamental set of solutions for the given equation, because they are linearly independent solutions.** --- ## **Problem 2** **Given:** \( Y_1(t) = e^{-2t} \) and \( Y_2(t) = e^{-3t} \) form a fundamental set for \( y'' + 5y' + 6y = 0 \). **Find:** The natural fundamental set of solutions associated with \( t = 0 \). --- ### **Step 1: Write the general solution and evaluate at \( t = 0 \)** The general solution is: \[ y(t) = c_1 e^{-2t} + c_2 e^{-3t} \] At \( t = 0 \): \[ y(0) = c_1 e^{0} + c_2 e^{0} = c_1 + c_2 \] \[ y'(t) = -2c_1 e^{-2t} - 3c_2 e^{-3t} \] \[ y'(0) = -2c_1 - 3c_2 \] **Explanation:** We are preparing to find initial conditions that correspond to the "natural" fundamental set. --- ### **Step 2: Find solutions corresponding to initial conditions** A natural fundamental set corresponds to: - \( y_1(0) = 1, y_1'(0) = 0 \) - \( y_2(0) = 0, y_2'(0) = 1 \) Let \( y_1(t) = a_1 e^{-2t} + a_2 e^{-3t} \), and set \( y_1(0) = 1, y_1'(0) = 0 \): \[ y_1(0):\quad a_1 + a_2 = 1 \] \[ y_1'(0):\quad -2a_1 - 3a_2 = 0 \] Solve: \[ a_1 + a_2 = 1 \\ -2a_1 - 3a_2 = 0 \implies -2a_1 - 3(1-a_1) = 0 \implies -2a_1 - 3 + 3a_1 = 0 \implies a_1 = 3, a_2 = -2 \] So, \( y_1(t) = 3e^{-2t} - 2e^{-3t} \) Now, for \( y_2(t) \) with \( y_2(0) = 0, y_2'(0) = 1 \): \[ b_1 + b_2 = 0 \\ -2b_1 - 3b_2 = 1 \] \[ b_1 + b_2 = 0 \implies b_2 = -b_1 \\ -2b_1 - 3(-b_1) = 1 \implies -2b_1 + 3b_1 = 1 \implies b_1 = 1, b_2 = -1 \] So, \( y_2(t) = e^{-2t} - e^{-3t} \) **Explanation:** We found the "natural" fundamental set associated with initial conditions at \( t = 0 \). --- ### **Final Answer for Problem 2:** **The natural fundamental set of solutions associated with \( t = 0 \) is:** \[ \boxed{ y_1(t) = 3e^{-2t} - 2e^{-3t}, \quad y_2(t) = e^{-2t} - e^{-3t} } \] --- **Summary Table:** | Problem | Step 1 | Step 2 | Final Answer | |---------|--------|--------|-------------| | 1 | Check if solutions | Check Wronskian | Yes, they form a fundamental set | | 2 | General solution at \( t=0 \) | Find natural set | \( y_1(t) = 3e^{-2t} - 2e^{-3t} \), \( y_2(t) = e^{-2t} - e^{-3t} \) | Let me know if you'd like additional clarifications!