Let's solve the given system of differential equations by systematic elimination. Given: \[ \begin{cases} D^2x - D y = t \\ (D+9)x + (D+9)y = 8 \end{cases} \] where \( D = \frac{d}{dt} \). Let's write them out explicitly: 1. \( x'' - y' = t \) 2. \( (D+9)x + (D+9)y = 8 \Rightarrow x' + 9x + y' + 9y = 8 \) Let's simplify equation (2): \[ x' + 9x + y' + 9y = 8 \implies (x' + y') + 9(x + y) = 8 \] \[ (x' + y') = \frac{d}{dt}(x + y) \] So, \[ \frac{d}{dt}(x + y) + 9(x + y) = 8 \] Let \( S = x + y \), then \[ S' + 9S = 8 \] This is a first order linear ODE for \( S \): ### Step 1: Solve for \( S \) The integrating factor (IF) is \( e^{9t} \): \[ e^{9t} S' + 9 e^{9t} S = 8 e^{9t} \] \[ \frac{d}{dt}(e^{9t} S) = 8 e^{9t} \] Integrate both sides: \[ e^{9t} S = \int 8 e^{9t} dt = \frac{8}{9} e^{9t} + C \] \[ S = x + y = \frac{8}{9} + Ce^{-9t} \] ### Step 2: Express \( y \) in terms of \( x \) \[ x + y = \frac{8}{9} + Ce^{-9t} \implies y = \frac{8}{9} + Ce^{-9t} - x \] ### Step 3: Substitute \( y \) into equation (1) Recall equation (1): \[ x'' - y' = t \] First, compute \( y' \): \[ y' = \frac{d}{dt} \left( \frac{8}{9} + Ce^{-9t} - x \right) = 0 - 9C e^{-9t} - x' = -9C e^{-9t} - x' \] Now, substitute into equation (1): \[ x'' - (-9C e^{-9t} - x') = t \implies x'' + x' + 9C e^{-9t} = t \] ### Step 4: Solve for \( x \) This is a non-homogeneous second order ODE: \[ x'' + x' = t - 9C e^{-9t} \] Let's solve the homogeneous part first: #### **Homogeneous equation:** \[ x'' + x' = 0 \] The characteristic equation: \[ m^2 + m = 0 \implies m(m+1) = 0 \implies m = 0, -1 \] So, \[ x_h(t) = A + B e^{-t} \] #### **Particular solution:** Let's look for a particular solution for the RHS \( t - 9C e^{-9t} \): - For \( t \): try \( x_{p1} = at + b \) - For \( -9C e^{-9t} \): try \( x_{p2} = k e^{-9t} \) First, for \( x_{p1} = at + b \): \[ x_{p1}' = a \\ x_{p1}'' = 0 \\ x_{p1}'' + x_{p1}' = a = t \implies a = t \textrm{?} \] But that doesn't make sense; let's try \( x_{p1} = p t + q \): \[ x_{p1}' = p \\ x_{p1}'' = 0 \\ x_{p1}'' + x_{p1}' = p = t \implies p = t \textrm{?} \] But \( t \) is not a constant. Let's try \( x_{p1} = \alpha t + \beta \): But then \( x_{p1}' = \alpha \), and \( x_{p1}'' = 0 \). So, \( x_{p1}'' + x_{p1}' = \alpha \), but \( t \) is on the right. Hence, we need to try polynomial of higher degree: \( x_{p1} = a t^2 + b t + c \): \[ x_{p1}' = 2a t + b \\ x_{p1}'' = 2a \\ x_{p1}'' + x_{p1}' = 2a + 2a t + b = t \] Set equal to \( t \): \[ 2a t + (2a + b) = t \] So, \( 2a = 1 \implies a = \frac{1}{2} \), and \( 2a + b = 0 \implies b = -2a = -1 \). So, \[ x_{p1} = \frac{1}{2} t^2 - t \] Now, for \( x_{p2} = k e^{-9t} \): \[ x_{p2}' = -9k e^{-9t} \\ x_{p2}'' = 81 k e^{-9t} \\ x_{p2}'' + x_{p2}' = (81k - 9k) e^{-9t} = 72k e^{-9t} \] Set equal to \( -9C e^{-9t} \): \[ 72k = -9C \implies k = -\frac{9C}{72} = -\frac{C}{8} \] So, \[ x_{p2} = -\frac{C}{8} e^{-9t} \] Combine all: \[ x(t) = A + B e^{-t} + \frac{1}{2} t^2 - t - \frac{C}{8} e^{-9t} \] ### Step 5: Recall \( y = \frac{8}{9} + Ce^{-9t} - x \): \[ y(t) = \frac{8}{9} + Ce^{-9t} - x(t) \] Plug in expression for \( x(t) \): \[ y(t) = \frac{8}{9} + Ce^{-9t} - \left[ A + B e^{-t} + \frac{1}{2} t^2 - t - \frac{C}{8} e^{-9t} \right] \] \[ = \frac{8}{9} - A - B e^{-t} - \frac{1}{2} t^2 + t + Ce^{-9t} + \frac{C}{8} e^{-9t} \] But \( Ce^{-9t} - \frac{C}{8} e^{-9t} = \frac{8C - C}{8} e^{-9t} = \frac{7C}{8} e^{-9t} \): So, \[ \boxed{ \begin{aligned} x(t) &= A + B e^{-t} + \frac{1}{2} t^2 - t - \frac{C}{8} e^{-9t} \\ y(t) &= \frac{8}{9} - A - B e^{-t} - \frac{1}{2} t^2 + t + \frac{7C}{8} e^{-9t} \end{aligned} } \] where \(A\), \(B\), and \(C\) are arbitrary constants determined by initial conditions (which are not provided). --- **Summary of Steps:** 1. Combine equations to get a single ODE in terms of \( x \). 2. Solve for \( x(t) \) including both homogeneous and particular solutions. 3. Express \( y(t) \) in terms of \( x(t) \). 4. Simplify to final general solution. --- **Final Answer:** \[ \boxed{ \begin{aligned} x(t) &= A + B e^{-t} + \frac{1}{2} t^2 - t - \frac{C}{8} e^{-9t} \\ y(t) &= \frac{8}{9} - A - B e^{-t} - \frac{1}{2} t^2 + t + \frac{7C}{8} e^{-9t} \end{aligned} } \]