Let's solve **Problem 1** step by step: ### Problem 1: \[ \begin{cases} x' = -x + 3y \\ y' = 2y \end{cases} \] --- ### **Step 1: Solve for \( y(t) \)** The second equation is **decoupled**: \[ y' = 2y \] This is a simple first-order linear ODE. **Solution:** \[ \frac{dy}{dt} = 2y \implies \frac{1}{y}dy = 2dt \implies \int \frac{1}{y}dy = \int 2dt \] \[ \ln|y| = 2t + C_1 \implies y = C_1 e^{2t} \] **Explanation:** We solved the equation for \( y \) using separation of variables and exponentiation. --- ### **Step 2: Substitute \( y(t) \) into the Equation for \( x(t) \)** Plug \( y = C_1 e^{2t} \) into the first equation: \[ x' = -x + 3y = -x + 3C_1 e^{2t} \] This is a nonhomogeneous first-order linear ODE for \( x \): \[ x' + x = 3C_1 e^{2t} \] The integrating factor is \( e^{\int 1 dt} = e^{t} \): Multiply both sides by \( e^t \): \[ e^t x' + e^t x = 3C_1 e^{3t} \implies \frac{d}{dt}(e^t x) = 3C_1 e^{3t} \] Integrate both sides: \[ e^t x = \int 3C_1 e^{3t} dt = 3C_1 \int e^{3t} dt = 3C_1 \frac{1}{3}e^{3t} + C_2 = C_1 e^{3t} + C_2 \] \[ x = C_1 e^{2t} + C_2 e^{-t} \] **Explanation:** We used the integrating factor method to solve for \( x(t) \). --- ### **Step 3: State the General Solution** Combine the results: \[ \boxed{ x(t) = C_1 e^{2t} + C_2 e^{-t},\quad y(t) = C_1 e^{2t} } \] **Explanation:** This is the general solution, where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions if given. --- ## **Final Answer:** \[ \boxed{ x(t) = C_1 e^{2t} + C_2 e^{-t},\quad y(t) = C_1 e^{2t} } \]

Let's solve **Problem 1** step-by-step, then briefly interpret the solution: --- ### **Step 1: Solve for \( y(t) \)** Given the system: \[ \begin{cases} x' = -x + 3y \\ y' = 2y \end{cases} \] Focus first on the second equation \( y' = 2y \), which is a straightforward first-order linear ODE. **Solution:** \[ \frac{dy}{dt} = 2y \] Separate variables: \[ \frac{dy}{y} = 2\,dt \] Integrate both sides: \[ \ln|y| = 2t + C_1 \] Exponentiate: \[ y(t) = C_1 e^{2t} \] **Explanation:** This step isolates \( y(t) \), which is essential because it simplifies the coupled system into a single differential equation for \( x(t) \). --- ### **Step 2: Solve for \( x(t) \) using the known \( y(t) \)** Substitute \( y(t) = C_1 e^{2t} \) into the first equation: \[ x' = -x + 3 y = -x + 3 C_1 e^{2t} \] This is a nonhomogeneous linear ODE: \[ x' + x = 3 C_1 e^{2t} \] The integrating factor \( \mu(t) \): \[ \mu(t) = e^{\int 1\,dt} = e^{t} \] Multiply through by \( e^{t} \): \[ e^{t} x' + e^{t} x = 3 C_1 e^{3t} \] Recognize the left side as the derivative of \( e^{t} x \): \[ \frac{d}{dt} \left( e^{t} x \right) = 3 C_1 e^{3t} \] Integrate both sides: \[ e^{t} x = \int 3 C_1 e^{3t} dt = 3 C_1 \cdot \frac{1}{3} e^{3t} + C_2 = C_1 e^{3t} + C_2 \] Solve for \( x(t) \): \[ x(t) = e^{-t} (C_1 e^{3t} + C_2) = C_1 e^{2t} + C_2 e^{-t} \] --- ### **Step 3: Write the general solution** \[ \boxed{ \begin{cases} x(t) = C_1 e^{2t} + C_2 e^{-t} \\ y(t) = C_1 e^{2t} \end{cases} } \] where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. --- ### **Final Answer:** The **general solution** of the system is: \[ \boxed{ x(t) = C_1 e^{2t} + C_2 e^{-t}, \quad y(t) = C_1 e^{2t} } \] --- **Interpretation:** The solutions show that \( y(t) \) depends solely on \( C_1 \), and \( x(t) \) involves two arbitrary constants, reflecting the system's general behavior. The exponential terms indicate growth or decay depending on initial conditions.