Let's carefully solve the given system of differential equations using the elimination method. The system is: \[ \begin{cases} 2x' + y' - 4x - 6y = e^{-3t} \quad (1) \\ x' + y' + 8x + 3y = e^{6t} \quad (2) \end{cases} \] --- ### **Step 1: Write the system in matrix form** The system involves derivatives \(x'\) and \(y'\). To facilitate elimination, first, let's write the equations explicitly: Equation (1): \(2x' + y' = 4x + 6y + e^{-3t}\) Equation (2): \(x' + y' = -8x - 3y + e^{6t}\) --- ### **Step 2: Eliminate \(y'\)** Subtract equation (2) from equation (1): \[ (2x' + y') - (x' + y') = (4x + 6y + e^{-3t}) - (-8x - 3y + e^{6t}) \] Simplify: \[ (2x' - x') + (y' - y') = 4x + 6y + e^{-3t} + 8x + 3y - e^{6t} \] \[ x' = (4x + 8x) + (6y + 3y) + e^{-3t} - e^{6t} \] \[ x' = 12x + 9y + e^{-3t} - e^{6t} \] **This is our first key equation:** \[ \boxed{ x' = 12x + 9y + e^{-3t} - e^{6t} } \] --- ### **Step 3: Express \(y'\) from equation (2)** From equation (2): \[ x' + y' = -8x - 3y + e^{6t} \] \[ \Rightarrow y' = -8x - 3y + e^{6t} - x' \] Substitute \(x'\) from the previous result: \[ y' = -8x - 3y + e^{6t} - [12x + 9y + e^{-3t} - e^{6t}] \] Simplify: \[ y' = -8x - 3y + e^{6t} - 12x - 9y - e^{-3t} + e^{6t} \] \[ y' = (-8x - 12x) + (-3y - 9y) + e^{6t} + e^{6t} - e^{-3t} \] \[ y' = -20x - 12y + 2e^{6t} - e^{-3t} \] --- ### **Step 4: Derive a second-order differential equation for \(x\)** Now, differentiate \(x'\) to get \(x''\): \[ x' = 12x + 9y + e^{-3t} - e^{6t} \] Differentiate both sides: \[ x'' = 12x' + 9 y' + \frac{d}{dt}(e^{-3t}) - \frac{d}{dt}(e^{6t}) \] Compute derivatives: \[ \frac{d}{dt}(e^{-3t}) = -3 e^{-3t} \] \[ \frac{d}{dt}(e^{6t}) = 6 e^{6t} \] Substitute: \[ x'' = 12x' + 9 y' - 3 e^{-3t} - 6 e^{6t} \] Now, substitute \(x'\) and \(y'\): \[ x' = 12x + 9 y + e^{-3t} - e^{6t} \] \[ y' = -20x - 12 y + 2 e^{6t} - e^{-3t} \] Plug into the expression for \(x''\): \[ x'' = 12(12x + 9 y + e^{-3t} - e^{6t}) + 9(-20x - 12 y + 2 e^{6t} - e^{-3t}) - 3 e^{-3t} - 6 e^{6t} \] Simplify term by term: \[ x'' = 144x + 108 y + 12 e^{-3t} - 12 e^{6t} - 180x - 108 y + 18 e^{6t} - 9 e^{-3t} - 3 e^{-3t} - 6 e^{6t} \] Combine like terms: - \(x\): \(144x - 180x = -36x\) - \(y\): \(108 y - 108 y = 0\) - \(e^{-3t}\): \(12 e^{-3t} - 9 e^{-3t} - 3 e^{-3t} = 0\) - \(e^{6t}\): \(-12 e^{6t} + 18 e^{6t} - 6 e^{6t} = 0\) Thus, all exponential terms cancel out! **Final second-order differential equation for \(x\):** \[ \boxed{ x'' = -36 x } \] --- ### **Step 5: Solve the second-order differential equation** This is a simple harmonic oscillator: \[ x'' + 36 x = 0 \] The characteristic equation: \[ r^2 + 36 = 0 \Rightarrow r^2 = -36 \Rightarrow r = \pm 6i \] General solution: \[ x(t) = C_1 \cos 6t + C_2 \sin 6t \] where \(C_1, C_2\) are arbitrary constants. --- ### **Step 6: Find \(y(t)\)** Recall from earlier: \[ x' = 12x + 9 y + e^{-3t} - e^{6t} \] Rearranged to solve for \(y\): \[ 9 y = x' - 12x - e^{-3t} + e^{6t} \] \[ y = \frac{1}{9} \left( x' - 12x - e^{-3t} + e^{6t} \right) \] Calculate \(x'\): \[ x' = -6 C_1 \sin 6t + 6 C_2 \cos 6t \] Now, substitute \(x(t)\) and \(x'(t)\): \[ x(t) = C_1 \cos 6t + C_2 \sin 6t \] \[ x' = -6 C_1 \sin 6t + 6 C_2 \cos 6t \] Plug into \(y(t)\): \[ y(t) = \frac{1}{9} \left( -6 C_1 \sin 6t + 6 C_2 \cos 6t - 12 (C_1 \cos 6t + C_2 \sin 6t) - e^{-3t} + e^{6t} \right) \] Simplify the terms: \[ y(t) = \frac{1}{9} \left( -6 C_1 \sin 6t + 6 C_2 \cos 6t - 12 C_1 \cos 6t - 12 C_2 \sin 6t - e^{-3t} + e^{6t} \right) \] Express as: \[ y(t) = \frac{1}{9} \left( \left( -6 C_1 \sin 6t - 12 C_2 \sin 6t \right) + \left( 6 C_2 \cos 6t - 12 C_1 \cos 6t \right) - e^{-3t} + e^{6t} \right) \] Factor where possible: \[ y(t) = \frac{1}{9} \left( -6 C_1 \sin 6t - 12 C_2 \sin 6t + 6 C_2 \cos 6t - 12 C_1 \cos 6t - e^{-3t} + e^{6t} \right) \] --- ### **### **Final General Solution:** \[ \boxed{ \begin{aligned} x(t) &= C_1 \cos 6t + C_2 \sin 6t \\ y(t) &= \frac{1}{9} \left[ -6 C_1 \sin 6t - 12 C_2 \sin 6t + 6 C_2 \cos 6t - 12 C_1 \cos 6t \right] + \frac{1}{9} \left( - e^{-3t} + e^{6t} \right) \end{aligned} } \] where \(C