1. **State the problem:** Solve the differential equation $$ (D^2 - 3D + 2)y = 2\cos(2x+3) + 2e^x $$ where $D = \frac{d}{dx}$. 2. **Identify the type of equation:** This is a nonhomogeneous linear differential equation with constant coefficients. 3. **Solve the homogeneous equation:** $$ (D^2 - 3D + 2)y = 0 $$ Characteristic equation: $$ r^2 - 3r + 2 = 0 $$ Factor: $$ (r - 1)(r - 2) = 0 $$ Roots: $$ r = 1, 2 $$ General solution to homogeneous: $$ y_h = C_1 e^x + C_2 e^{2x} $$ 4. **Find a particular solution $y_p$:** The right side is: $$ 2\cos(2x+3) + 2e^x $$ We find particular solutions for each term separately. - For $$ 2e^x $$, since $e^x$ is a solution to the homogeneous equation (root $r=1$), multiply by $x$: Try: $$ y_{p1} = A x e^x $$ - For $$ 2\cos(2x+3) $$, try: $$ y_{p2} = M \cos(2x+3) + N \sin(2x+3) $$ 5. **Calculate derivatives for $y_{p1}$:** $$ y_{p1} = A x e^x $$ $$ y_{p1}' = A e^x + A x e^x $$ $$ y_{p1}'' = A e^x + A e^x + A x e^x = 2 A e^x + A x e^x $$ Apply operator: $$ (D^2 - 3D + 2) y_{p1} = y_{p1}'' - 3 y_{p1}' + 2 y_{p1} $$ Substitute: $$ = (2 A e^x + A x e^x) - 3 (A e^x + A x e^x) + 2 (A x e^x) $$ $$ = 2 A e^x + A x e^x - 3 A e^x - 3 A x e^x + 2 A x e^x $$ $$ = (2 A e^x - 3 A e^x) + (A x e^x - 3 A x e^x + 2 A x e^x) $$ $$ = - A e^x + 0 $$ $$ = - A e^x $$ Set equal to $$ 2 e^x $$: $$ - A e^x = 2 e^x \implies A = -2 $$ 6. **Calculate derivatives for $y_{p2}$:** $$ y_{p2} = M \cos(2x+3) + N \sin(2x+3) $$ $$ y_{p2}' = -2 M \sin(2x+3) + 2 N \cos(2x+3) $$ $$ y_{p2}'' = -4 M \cos(2x+3) - 4 N \sin(2x+3) $$ Apply operator: $$ (D^2 - 3D + 2) y_{p2} = y_{p2}'' - 3 y_{p2}' + 2 y_{p2} $$ Substitute: $$ = (-4 M \cos(2x+3) - 4 N \sin(2x+3)) - 3 (-2 M \sin(2x+3) + 2 N \cos(2x+3)) + 2 (M \cos(2x+3) + N \sin(2x+3)) $$ Group terms: $$ = (-4 M + 2 M) \cos(2x+3) + (-4 N + 2 N) \sin(2x+3) + 6 M \sin(2x+3) - 6 N \cos(2x+3) $$ $$ = (-2 M - 6 N) \cos(2x+3) + (-2 N + 6 M) \sin(2x+3) $$ Set equal to $$ 2 \cos(2x+3) + 0 \sin(2x+3) $$: System: $$ -2 M - 6 N = 2 $$ $$ -2 N + 6 M = 0 $$ From second: $$ -2 N + 6 M = 0 \implies N = 3 M $$ Substitute into first: $$ -2 M - 6 (3 M) = 2 \implies -2 M - 18 M = 2 \implies -20 M = 2 \implies M = -\frac{1}{10} $$ Then: $$ N = 3 \times -\frac{1}{10} = -\frac{3}{10} $$ 7. **Write the particular solution:** $$ y_p = y_{p1} + y_{p2} = -2 x e^x - \frac{1}{10} \cos(2x+3) - \frac{3}{10} \sin(2x+3) $$ 8. **Write the general solution:** $$ y = y_h + y_p = C_1 e^x + C_2 e^{2x} - 2 x e^x - \frac{1}{10} \cos(2x+3) - \frac{3}{10} \sin(2x+3) $$