Let's solve the given ODE step by step. The system is: \[ x'(t) = -4x - 3y - 5 \] Since only one equation is provided, we assume either \( y(t) \) is given, or the full system has more equations (but only one is shown). Let's solve for \( x(t) \) **in terms of \( y(t) \)**. --- ### **Step 1: Express the equation in standard linear ODE form** \[ x'(t) + 4x + 3y = -5 \] This is a first-order linear ODE for \( x(t) \), with \( y(t) \) as a "forcing" (nonhomogeneous) term. **Explanation:** We wrote the equation in the standard form \( x' + P(t)x = Q(t) \) to prepare for integrating factor method. --- ### **Step 2: Find the integrating factor and solve for \( x(t) \)** The integrating factor (IF) is: \[ \mu(t) = e^{\int 4 dt} = e^{4t} \] Now, multiply both sides by the integrating factor: \[ e^{4t} x'(t) + 4 e^{4t} x(t) + 3 e^{4t} y(t) = -5 e^{4t} \] The left side is the derivative: \[ \frac{d}{dt}[e^{4t} x(t)] + 3 e^{4t} y(t) = -5 e^{4t} \] Integrate both sides: \[ e^{4t} x(t) = \int [-3 e^{4t} y(t) - 5 e^{4t}] dt + C \] So, \[ x(t) = e^{-4t} \left( \int [-3 e^{4t} y(t) - 5 e^{4t}] dt + C \right) \] **Explanation:** We applied the integrating factor method to solve the nonhomogeneous linear ODE. The solution is expressed in terms of \( y(t) \). --- ## **Final Answer** \[ \boxed{ x(t) = e^{-4t} \left( \int [-3 e^{4t} y(t) - 5 e^{4t}] dt + C \right) } \] This gives the general solution for \( x(t) \) **in terms of \( y(t) \)**. To fully solve the system, you would need a second equation for \( y(t) \).