Let's solve problem 1 as requested, in **3 steps** with an explanation at each step, and the final answer at the end: ### Problem 1 \[ x' = -x + 3y, \quad y' = 2y \] --- ## **Step 1: Solve for \( y(t) \) independently** Notice \( y' = 2y \) is a first-order linear ODE for \( y \): \[ y' = 2y \] This is separable. Divide both sides by \( y \) (assuming \( y \neq 0 \)): \[ \frac{dy}{dt} = 2y \implies \frac{dy}{y} = 2\,dt \] Integrate both sides: \[ \ln|y| = 2t + C_1 \implies y(t) = C_1 e^{2t} \] **Explanation:** We solved the second equation first because it's independent of \( x \). The solution is an exponential function. --- ## **Step 2: Substitute \( y(t) \) into the equation for \( x(t) \)** Now use \( y(t) = C_1 e^{2t} \) in the equation for \( x \): \[ x' = -x + 3y = -x + 3C_1 e^{2t} \] This is a non-homogeneous first-order linear ODE for \( x \): \[ x' + x = 3C_1 e^{2t} \] The integrating factor is \( e^{t} \): Multiply both sides by \( e^{t} \): \[ e^{t} x' + e^{t} x = 3C_1 e^{2t} e^{t} \implies \frac{d}{dt}(e^{t} x) = 3C_1 e^{3t} \] Integrate both sides: \[ e^{t} x = \int 3C_1 e^{3t} dt = 3C_1 \int e^{3t} dt = 3C_1 \frac{1}{3} e^{3t} + C_2 = C_1 e^{3t} + C_2 \] So, \[ x(t) = C_1 e^{2t} + C_2 e^{-t} \] **Explanation:** We substituted the previously found \( y(t) \) into the first equation and solved the resulting linear ODE using an integrating factor. --- ## **Step 3: Write the general solution** Recall the solutions: \[ y(t) = C_1 e^{2t} \] \[ x(t) = C_1 e^{2t} + C_2 e^{-t} \] **Explanation:** The general solution is found by expressing both variables in terms of arbitrary constants \( C_1 \) and \( C_2 \), representing the family of solutions. --- ## **Final Answer:** \[ \boxed{ x(t) = C_1 e^{2t} + C_2 e^{-t}, \quad y(t) = C_1 e^{2t} } \] where \( C_1, C_2 \) are arbitrary constants.

Let's proceed step-by-step to solve the system: \[ x' = -x + 3y \quad \text{and} \quad y' = 2y \] --- ### **Step 1: Solve for \( y(t) \) independently** **Explanation:** The second equation \( y' = 2y \) is a simple first-order linear ODE. Solving it yields the general form of \( y(t) \). \[ \text{Separate variables:} \quad \frac{dy}{dt} = 2y \] \[ \Rightarrow \frac{dy}{y} = 2\,dt \] \[ \Rightarrow \ln|y| = 2t + C_1 \] \[ \Rightarrow y(t) = C_1 e^{2t} \] **Result:** The solution for \( y(t) \) is: \[ \boxed{ y(t) = C_1 e^{2t} } \] --- ### **Step 2: Substitute \( y(t) \) into the \( x \) equation and solve** **Explanation:** Plugging \( y(t) = C_1 e^{2t} \) into the first equation: \[ x' = -x + 3 y(t) = -x + 3 C_1 e^{2t} \] This is a linear first-order ODE in \( x(t) \): \[ x' + x = 3 C_1 e^{2t} \] The integrating factor (IF) is: \[ \mu(t) = e^{\int 1\,dt} = e^{t} \] Multiply through by \( e^{t} \): \[ e^{t} x' + e^{t} x = 3 C_1 e^{2t} e^{t} = 3 C_1 e^{3t} \] Observe that: \[ \frac{d}{dt} \left( e^{t} x \right) = 3 C_1 e^{3t} \] Integrate both sides: \[ e^{t} x = \int 3 C_1 e^{3t} dt = 3 C_1 \frac{1}{3} e^{3t} + C_2 = C_1 e^{3t} + C_2 \] Solve for \( x(t) \): \[ x(t) = e^{-t} \left( C_1 e^{3t} + C_2 \right) = C_1 e^{2t} + C_2 e^{-t} \] --- ### **Step 3: Write the **general solution** of the system** **Explanation:** Combining the results: \[ \boxed{ \begin{cases} x(t) = C_1 e^{2t} + C_2 e^{-t} \\ y(t) = C_1 e^{2t} \end{cases} } \] where \( C_1, C_2 \) are arbitrary constants determined by initial conditions. --- ## **Final answer:** \[ \boxed{ \begin{aligned} x(t) &= C_1 e^{2t} + C_2 e^{-t} \\ y(t) &= C_1 e^{2t} \end{aligned} } \] This represents the **general solution** to the system.