To find the general solution of the differential equation \( y'' - 2y' - 3y = -3te^{-t} \), we need to solve both the homogeneous and particular solutions. ### Step 1: Solve the Homogeneous EquationThe homogeneous equation is: \[ y'' - 2y' - 3y = 0 \] 1. **Find the characteristic equation:** \[ r^2 - 2r - 3 = 0 \] 2. **Solve the characteristic equation:** \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ r = \frac{2 \pm \sqrt{4 + 12}}{2} \] \[ r = \frac{2 \pm \sqrt{16}}{2} \] \[ r = \frac{2 \pm 4}{2} \] \[ r_1 = 3, \quad r_2 = -1 \] 3. **Write the general solution of the homogeneous equation:** \[ y_h(t) = c_1 e^{3t} + c_2 e^{-t} \] ### Step 2: Find the Particular SolutionWe need to find a particular solution \( y_p(t) \) for the non-homogeneous equation: \[ y'' - 2y' - 3y = -3te^{-t} \] Since the right-hand side is \( -3te^{-t} \), we try a particular solution of the form: \[ y_p(t) = (At + B)e^{-t} \] 1. **Compute the derivatives:** \[ y_p'(t) = A e^{-t} - (At + B)e^{-t} = (A - At - B)e^{-t} \] \[ y_p''(t) = -A e^{-t} - (A - At - B)e^{-t} = (-A - A + At + B)e^{-t} = (-2A + At + B)e^{-t} \] 2. **Substitute into the differential equation:** \[ (-2A + At + B)e^{-t} - 2(A - At - B)e^{-t} - 3(At + B)e^{-t} = -3te^{-t} \] \[ (-2A + At + B) - 2A + 2At + 2B - 3At - 3B = -3t \] \[ (-2A - 2A + B + 2B - 3B) + (At + 2At - 3At) = -3t \] \[ -4A - B + 0t = -3t \] 3. **Match coefficients:** For the \( t \) term: \[ 0 = -3 \] This is not possible, so we need to adjust our particular solution form. Let's try: \[ y_p(t) = (At^2 + Bt)e^{-t} \] 4. **Compute the derivatives:** \[ y_p'(t) = (2At + B)e^{-t} - (At^2 + Bt)e^{-t} = (2At + B - At^2 - Bt)e^{-t} = (-At^2 + At)e^{-t} \] \[ y_p''(t) = (-2A)e^{-t} - (2At + B)e^{-t} = (-2A - 2At - B)e^{-t} \] 5. **Substitute into the differential equation:** \[ (-2A - 2At - B)e^{-t} - 2(2At + B)e^{-t} - 3(At^2 + Bt)e^{-t} = -3te^{-t} \] \[ (-2A - 2At - B) - 4At - 2B - 3At^2 - 3Bt = -3t \] \[ -3At^2 - 6At - 3B - 2A = -3t \] 6. **Match coefficients:** For the \( t^2 \) term: \[ -3A = 0 \] \[ A = 0 \] For the \( t \) term: \[ -6A - 3B = -3 \] \[ -3B = -3 \] \[ B = 1 \] For the constant

To find the general solution of the differential equation $y'' - 2y' - 3y = -3te^{-t}$, we need to solve both the homogeneous and particular solutions. ### Step 1: Solve the Homogeneous EquationThe homogeneous equation is: $y'' - 2y' - 3y = 0$ 1. Find the characteristic equation: $r^2 - 2r - 3 = 0$ 2. Solve the characteristic equation: $r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}$ $r = \frac{2 \pm \sqrt{4 + 12}}{2}$ $r = \frac{2 \pm \sqrt{16}}{2}$ $r = \frac{2 \pm 4}{2}$ $r_1 = 3, \quad r_2 = -1$ 3. Write the general solution of the homogeneous equation: $y_h(t) = c_1 e^{3t} + c_2 e^{-t}$ ### Step 2: Find the Particular SolutionWe need to find a particular solution $y_p(t)$ for the non-homogeneous equation: $y'' - 2y' - 3y = -3te^{-t}$ Since the right-hand side is $-3te^{-t}$, we try a particular solution of the form: $y_p(t) = (At + B)e^{-t}$ 1. Compute the derivatives: $y_p'(t) = A e^{-t} - (At + B)e^{-t} = (A - At - B)e^{-t}$ $y_p''(t) = -A e^{-t} - (A - At - B)e^{-t} = (-A - A + At + B)e^{-t} = (-2A + At + B)e^{-t}$ 2. Substitute into the differential equation: $(-2A + At + B)e^{-t} - 2(A - At - B)e^{-t} - 3(At + B)e^{-t} = -3te^{-t}$ $(-2A + At + B) - 2A + 2At + 2B - 3At - 3B = -3t$ $(-2A - 2A + B + 2B - 3B) + (At + 2At - 3At) = -3t$ $-4A - B + 0t = -3t$ 3. Match coefficients: For the $t$ term: $0 = -3$ This is not possible, so we need to adjust our particular solution form. Let's try: $y_p(t) = (At^2 + Bt)e^{-t}$ 4. Compute the derivatives: $y_p'(t) = (2At + B)e^{-t} - (At^2 + Bt)e^{-t} = (2At + B - At^2 - Bt)e^{-t} = (-At^2 + At)e^{-t}$ $y_p''(t) = (-2A)e^{-t} - (2At + B)e^{-t} = (-2A - 2At - B)e^{-t}$ 5. Substitute into the differential equation: $(-2A - 2At - B)e^{-t} - 2(2At + B)e^{-t} - 3(At^2 + Bt)e^{-t} = -3te^{-t}$ $(-2A - 2At - B) - 4At - 2B - 3At^2 - 3Bt = -3t$ $-3At^2 - 6At - 3B - 2A = -3t$ 6. Match coefficients: For the $t^2$ term: $-3A = 0$ $A = 0$ For the $t$ term: $-6A - 3B = -3$ $-3B = -3$ $B = 1$ For the constant