Let's clarify the system from the image: Given: \[ \begin{cases} \dot{x} = y \\ \dot{y} = 2x + 3y - 2 \end{cases} \] Let's solve this in 2 steps with explanations: --- ### **Step 1: Combine into a single second-order ODE** We have: - \(\dot{x} = y\) - \(\dot{y} = 2x + 3y - 2\) Differentiate the first equation with respect to \(t\): \[ \frac{d}{dt}(\dot{x}) = \frac{d}{dt}(y) \implies \ddot{x} = \dot{y} \] Now, substitute \(\dot{y}\) from the second equation: \[ \ddot{x} = 2x + 3y - 2 \] But from the first equation, \(y = \dot{x}\): \[ \ddot{x} = 2x + 3\dot{x} - 2 \] **Explanation:** We have reduced the system to a single second-order ODE: \[ \ddot{x} - 3\dot{x} - 2x = -2 \] --- ### **Step 2: Solve the second-order ODE** The equation is: \[ \ddot{x} - 3\dot{x} - 2x = -2 \] **Homogeneous solution:** Solve \(\ddot{x} - 3\dot{x} - 2x = 0\) The characteristic equation: \[ r^2 - 3r - 2 = 0 \] \[ r = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \] So, homogeneous solution: \[ x_h(t) = A e^{r_1 t} + B e^{r_2 t} \] where \(r_1 = \frac{3 + \sqrt{17}}{2}\), \(r_2 = \frac{3 - \sqrt{17}}{2}\). **Particular solution:** For the constant right side, try \(x_p = K\): Plug into the ODE: \[ 0 - 0 - 2K = -2 \implies K = 1 \] **Total solution:** \[ x(t) = A e^{r_1 t} + B e^{r_2 t} + 1 \] Recall \(y = \dot{x}\): \[ y(t) = A r_1 e^{r_1 t} + B r_2 e^{r_2 t} \] **Explanation:** We solved the second-order ODE, determined the complementary and particular solutions, and expressed both \(x(t)\) and \(y(t)\) in terms of exponentials. --- ## **Final Answer:** \[ \boxed{ \begin{aligned} x(t) &= A e^{r_1 t} + B e^{r_2 t} + 1 \\ y(t) &= A r_1 e^{r_1 t} + B r_2 e^{r_2 t} \end{aligned} } \] where \(r_1 = \frac{3 + \sqrt{17}}{2}\), \(r_2 = \frac{3 - \sqrt{17}}{2}\), and \(A, B\) are constants determined by initial conditions.

Let's clarify the problem: You provided the relation: \[ y(t) = 2x + 3y - 2 \] However, this appears to be an equation involving \(x(t)\) and \(y(t)\). To proceed, I will assume that the system of ODEs you're referring to is: \[ \begin{cases} \dot{x} = y \\ \dot{y} = 2x + 3y - 2 \end{cases} \] which aligns with similar previous problems, as the relation \( y(t) = 2x + 3y - 2 \) can be rearranged or used as a differential relation. Alternatively, if your statement was a typo or shorthand for the system, the solution approach is similar. --- ### **Step 1: Convert the system into a single second-order ODE** Given: \[ \dot{x} = y \] \[ \dot{y} = 2x + 3y - 2 \] Differentiate \(\dot{x} = y\): \[ \ddot{x} = \dot{y} \] Substitute \(\dot{y}\): \[ \ddot{x} = 2x + 3y - 2 \] Since \(y = \dot{x}\): \[ \ddot{x} = 2x + 3 \dot{x} - 2 \] **This leads to the second-order ODE:** \[ \boxed{ \ddot{x} - 3 \dot{x} - 2x = -2 } \] **Explanation:** By differentiating the first equation and substituting from the second, we reduce the system to a single second-order linear nonhomogeneous differential equation in \(x(t)\). --- ### **Step 2: Solve the second-order ODE** The equation: \[ \ddot{x} - 3 \dot{x} - 2x = -2 \] **Homogeneous part:** \[ \ddot{x} - 3 \dot{x} - 2x = 0 \] Characteristic equation: \[ r^2 - 3r - 2 = 0 \] Solve for \(r\): \[ r = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \] Homogeneous solution: \[ x_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] where \[ r_1 = \frac{3 + \sqrt{17}}{2}, \quad r_2 = \frac{3 - \sqrt{17}}{2} \] **Particular solution:** Since RHS is a constant \(-2\), try \(x_p = K\): Plug into the ODE: \[ 0 - 0 - 2K = -2 \implies K = 1 \] **General solution:** \[ x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + 1 \] Recall \(y(t) = \dot{x}(t)\): \[ y(t) = C_1 r_1 e^{r_1 t} + C_2 r_2 e^{r_2 t} \] --- ## **Final answer:** \[ \boxed{ \begin{aligned} x(t) &= C_1 e^{\frac{3 + \sqrt{17}}{2} t} + C_2 e^{\frac{3 - \sqrt{17}}{2} t} + 1 \\ y(t) &= C_1 \left(\frac{3 + \sqrt{17}}{2}\right) e^{\frac{3 + \sqrt{17}}{2} t} + C_2 \left(\frac{3 - \sqrt{17}}{2}\right) e^{\frac{3 - \sqrt{17}}{2} t} \end{aligned} } \] where \(C_1, C_2\) are arbitrary constants determined by initial conditions. --- **Note:** If the original statement was different or involved additional relations, please clarify for a more precise solution.