1. **Problem:** Solve the differential equation $(D^2 + D)y = -\cos x$ using the method of undetermined coefficients. Step 1: Identify the characteristic equation for the complementary solution (C.S.): $$D^2 + D = 0 \implies r^2 + r = 0 \implies r(r+1) = 0\implies r=0,-1.$$ Step 2: Write the complementary solution: $$y_c = C_1 + C_2 e^{-x}.$$ Step 3: Find a particular solution $y_p$. Since the right side is $-\cos x$, try: $$y_p = A \cos x + B \sin x.$$ Step 4: Compute derivatives: $$y_p' = -A \sin x + B \cos x, \quad y_p'' = -A \cos x - B \sin x.$$ Step 5: Apply the differential operator: $$D^2 y_p + D y_p = y_p'' + y_p' = (-A \cos x - B \sin x) + (-A \sin x + B \cos x) = (-A + B) \cos x + (-B - A) \sin x.$$ Step 6: Set equal to $-\cos x$: $$(-A + B) \cos x + (-B - A) \sin x = -\cos x + 0 \sin x.$$ Equate coefficients: $$-A + B = -1, \quad -B - A = 0.$$ Step 7: Solve system: From second, $-B - A=0 \implies B=-A$. Substitute into first: $$-A + (-A) = -1 \implies -2A = -1 \implies A = \frac{1}{2}.$$ Thus, $B = -\frac{1}{2}$. Step 8: Write final solution: $$y = y_c + y_p = C_1 + C_2 e^{-x} + \frac{1}{2} \cos x - \frac{1}{2} \sin x.$$ 2. **Problem:** Solve $(D^2 - 6D + 9)y = e^x$. Step 1: Characteristic equation: $$r^2 - 6r + 9=0 \implies (r-3)^2=0.$$ Step 2: Complementary solution: $$y_c = (C_1 + C_2 x) e^{3x}.$$ Step 3: Particular solution guess because RHS is $e^{x}$, and $e^{x}$ is not a root of the characteristic equation: $$y_p = A e^{x}.$$ Step 4: Compute: $$y_p' = A e^x, \, y_p'' = A e^x.$$ Apply operator: $$(D^2 - 6D + 9)y_p = y_p'' - 6 y_p' + 9 y_p = A e^x - 6 A e^x + 9 A e^x = (1 - 6 + 9) A e^x = 4 A e^x.$$ Step 5: Set equal to RHS: $$4 A e^x = e^x \implies 4A = 1 \implies A = \frac{1}{4}.$$ Step 6: Final solution: $$y = (C_1 + C_2 x) e^{3x} + \frac{1}{4} e^{x}.$$ 3. **Problem:** Solve $(D^2 + 3D + 2)y = 12 x^2$. Step 1: Characteristic equation: $$r^2 + 3r + 2 = 0 \implies (r+1)(r+2)=0 \implies r=-1,-2.$$ Step 2: Complementary solution: $$y_c = C_1 e^{-x} + C_2 e^{-2x}.$$ Step 3: RHS is polynomial of degree 2, guess: $$y_p = A x^2 + B x + C.$$ Step 4: Compute derivatives: $$y_p' = 2 A x + B, \quad y_p'' = 2 A.$$ Step 5: Apply operator: $$(D^2 + 3D + 2) y_p = y_p'' + 3 y_p' + 2 y_p = 2 A + 3 (2 A x + B) + 2 (A x^2 + B x + C) = 2A + 6 A x + 3 B + 2 A x^2 + 2 B x + 2 C.$$ Step 6: Group powers of $x$: $$2 A x^2 + (6 A + 2 B) x + (2 A + 3 B + 2 C) = 12 x^2 + 0 x + 0.$$ Step 7: Equate coefficients: $$2A = 12 \implies A = 6,$$ $$6A + 2B = 0 \implies 6(6) + 2B = 0 \implies 36 + 2B = 0 \implies B = -18,$$ $$2A + 3B + 2C = 0 \implies 2(6) + 3(-18) + 2C = 0 \implies 12 - 54 + 2C=0 \implies 2C=42 \implies C=21.$$ Step 8: Particular solution: $$y_p = 6 x^2 - 18 x + 21.$$ Step 9: Final solution: $$y = C_1 e^{-x} + C_2 e^{-2x} + 6 x^2 - 18 x + 21.$$ 4. **Problem:** Solve $(D^2 + 3D + 2)y = 1 + 3x + x^2$. Step 1: Characteristic equation same as above with roots $-1$ and $-2$. Step 2: Complementary solution: $$y_c = C_1 e^{-x} + C_2 e^{-2x}.$$ Step 3: RHS is polynomial degree 2, try: $$y_p = A x^2 + B x + C.$$ Step 4: Derivatives and operator apply as before: $$(D^2 + 3D + 2) y_p = 2 A x^2 + (6 A + 2 B) x + (2 A + 3 B + 2 C).$$ Step 5: Equate to RHS $1 + 3x + x^2$: $$2 A = 1 \implies A = \frac{1}{2},$$ $$6 A + 2 B = 3 \implies 6 \times \frac{1}{2} + 2 B = 3 \implies 3 + 2 B = 3 \implies 2 B = 0 \implies B=0,$$ $$2 A + 3 B + 2 C = 1 \implies 2 \times \frac{1}{2} + 0 + 2 C = 1 \implies 1 + 2 C = 1 \implies 2 C = 0 \implies C=0.$$ Step 6: Particular solution: $$y_p = \frac{1}{2} x^2 + 0 x + 0 = \frac{1}{2} x^2.$$ Step 7: Final solution: $$y = C_1 e^{-x} + C_2 e^{-2x} + \frac{1}{2} x^2.$$ 5. **Problem 7:** Solve $y'' - 3 y' -4 y = 30 e^x$. Step 1: Characteristic equation: $$r^2 - 3 r - 4=0 \implies (r-4)(r+1)=0 \implies r=4, -1.$$ Step 2: Complementary solution: $$y_c = C_1 e^{4 x} + C_2 e^{-x}.$$ Step 3: Particular solution guess: Since RHS is $30 e^{x}$ and $r=1$ is not root, try: $$y_p = A e^{x}.$$ Step 4: Compute operator on $y_p$: $$y_p' = A e^x, y_p''=A e^x,$$ $$y_p'' - 3 y_p' - 4 y_p = A e^x - 3 A e^x - 4 A e^x = (1 -3 -4) A e^x = -6 A e^x.$$ Step 5: Set equal to RHS: $$-6 A e^x = 30 e^x \implies -6 A = 30 \implies A = -5.$$ Step 6: Final solution: $$y = C_1 e^{4 x} + C_2 e^{-x} - 5 e^{x}.$$ 6. **Problem 8:** Solve $y'' - 3 y' -4 y = 30 e^{4 x}$. Step 1: Characteristic roots same as above. Step 2: RHS is $30 e^{4 x}$, but note $r=4$ is root of multiplicity 1. Step 3: Particular solution guess with multiplicity 1 root: $$y_p = A x e^{4 x}.$$ Step 4: Compute: $$y_p' = A e^{4 x} + 4 A x e^{4 x} = A (1 + 4 x) e^{4 x},$$ $$y_p'' = A (4 e^{4 x} + 4 e^{4 x} + 16 x e^{4 x}) = A (8 + 16 x) e^{4 x}.$$ Step 5: Apply operator: $$y_p'' - 3 y_p' - 4 y_p = A (8 + 16 x) e^{4 x} - 3 A (1 + 4 x) e^{4 x} - 4 A x e^{4 x} = e^{4 x} [A(8 + 16 x) - 3 A (1 + 4 x) - 4 A x].$$ Step 6: Simplify inside bracket: $$8 A + 16 A x - 3 A - 12 A x - 4 A x = (8 A - 3 A) + (16 A x - 12 A x - 4 A x) = 5 A + 0 = 5 A.$$ Step 7: Set equal to RHS: $$5 A e^{4 x} = 30 e^{4 x} \implies 5 A = 30 \implies A = 6.$$ Step 8: Final solution: $$y = C_1 e^{4 x} + C_2 e^{-x} + 6 x e^{4 x}.$$ 7. **Problem 11:** Solve $y'' - 4 y' + 3 y = 20 \cos x$. Step 1: Characteristic equation: $$r^2 - 4 r + 3=0 \implies (r - 3)(r -1) = 0 \implies r=3,1.$$ Step 2: Complementary solution: $$y_c = C_1 e^{3 x} + C_2 e^{x}.$$ Step 3: RHS is $20 \cos x$, try: $$y_p = A \cos x + B \sin x.$$ Step 4: Compute derivatives: $$y_p' = - A \sin x + B \cos x,$$ $$y_p'' = - A \cos x - B \sin x.$$ Step 5: Apply operator: $$y_p'' - 4 y_p' + 3 y_p = (- A \cos x - B \sin x) - 4 (- A \sin x + B \cos x) + 3 (A \cos x + B \sin x).$$ Step 6: Multiply and collect terms: $$(- A + 4 B + 3 A) \cos x + (- B + 4 A + 3 B) \sin x = (2 A + 4 B) \cos x + (4 A + 2 B) \sin x.$$ Step 7: Equate to RHS $20 \cos x + 0 \sin x$: $$2 A + 4 B = 20,$$ $$4 A + 2 B = 0.$$ Step 8: Solve system: From second: $4 A + 2 B = 0 \implies 2 A + B = 0 \implies B = -2 A.$ Substitute into first: $$2 A + 4 (-2 A) = 20 \implies 2 A -8 A = 20 \implies -6 A = 20 \implies A = - \frac{10}{3}.$$ Then $$B = -2 A = -2 (-\frac{10}{3}) = \frac{20}{3}.$$ Step 9: Final solution: $$y = C_1 e^{3 x} + C_2 e^{x} - \frac{10}{3} \cos x + \frac{20}{3} \sin x.$$ 8. **Problem 13:** Solve $y'' + 2 y' + y = 7 + 75 \sin 2 x$. Step 1: Characteristic equation: $$r^2 + 2 r + 1 = (r+1)^2 =0,$$ root $r = -1$ multiplicity 2. Step 2: Complementary solution: $$y_c = (C_1 + C_2 x) e^{-x}.$$ Step 3: RHS contains constant and sine, so try particular solution as sum: $$y_p = A + (B \cos 2x + C \sin 2x).$$ Step 4: Compute derivatives of $y_p$: $$y_p' = -2 B \sin 2 x + 2 C \cos 2 x,$$ $$y_p'' = -4 B \cos 2 x - 4 C \sin 2 x.$$ Step 5: Apply operator: $$y_p'' + 2 y_p' + y_p = [-4 B \cos 2 x - 4 C \sin 2 x] + 2 [-2 B \sin 2 x + 2 C \cos 2 x] + [A + B \cos 2 x + C \sin 2 x].$$ Step 6: Group terms: $$\cos 2 x: (-4 B + 4 C + B) = (-3 B + 4 C),$$ $$\sin 2 x: (-4 C - 4 B + C) = (-3 C - 4 B),$$ Constant term: $A$. Step 7: Set equal to RHS $7 + 75 \sin 2 x$: $$A = 7,$$ $$-3 B + 4 C = 0,$$ $$-3 C - 4 B = 75.$$ Step 8: Solve system for $B$ and $C$: From first: $$-3 B + 4 C = 0 \implies 4 C = 3 B \implies C = \frac{3}{4} B.$$ Substitute into second: $$-3 \left(\frac{3}{4} B \right) - 4 B = 75 \implies - \frac{9}{4} B - 4 B =75 \implies - \frac{9}{4} B - \frac{16}{4} B = 75 \implies - \frac{25}{4} B = 75 \implies B = - 12.$$ Then $$C = \frac{3}{4} (-12) = -9.$$ Step 9: Final solution: $$ y = (C_1 + C_2 x) e^{-x} + 7 - 12 \cos 2 x - 9 \sin 2 x.$$