1. **Problem Statement:** Solve the differential equation $$y'' - 4y = (x^2 - 3) \sin 2x$$ using the method of undetermined coefficients. 2. **General Approach:** The equation is linear with constant coefficients. The general solution is $$y = y_c + y_p$$ where $$y_c$$ is the complementary (homogeneous) solution and $$y_p$$ is a particular solution. 3. **Find the complementary solution $$y_c$$:** Solve the homogeneous equation: $$y'' - 4y = 0$$ The characteristic equation is: $$r^2 - 4 = 0$$ which factors as: $$ (r - 2)(r + 2) = 0$$ So, $$r = 2, -2$$ Therefore, $$y_c = C_1 e^{2x} + C_2 e^{-2x}$$ 4. **Form of the particular solution $$y_p$$:** The right side is $$ (x^2 - 3) \sin 2x $$, a product of a polynomial and sine function. For RHS of the form $$P_n(x) \sin(ax)$$ or $$P_n(x) \cos(ax)$$, try a particular solution of the form: $$y_p = (A x^2 + B x + C) \cos 2x + (D x^2 + E x + F) \sin 2x$$ where $$A,B,C,D,E,F$$ are constants to be determined. 5. **Calculate derivatives:** First derivative: $$y_p' = \frac{d}{dx} \left[(A x^2 + B x + C) \cos 2x + (D x^2 + E x + F) \sin 2x\right]$$ Use product rule and chain rule: $$y_p' = (2 A x + B) \cos 2x - 2 (A x^2 + B x + C) \sin 2x + (2 D x + E) \sin 2x + 2 (D x^2 + E x + F) \cos 2x$$ Simplify: $$y_p' = \left[(2 A x + B) + 2 (D x^2 + E x + F)\right] \cos 2x + \left[(2 D x + E) - 2 (A x^2 + B x + C)\right] \sin 2x$$ Second derivative: Differentiate $$y_p'$$ similarly: $$y_p'' = \frac{d}{dx} y_p'$$ Calculate derivatives of each term: For $$\cos 2x$$ term: $$\frac{d}{dx} \left[(2 A x + B) + 2 (D x^2 + E x + F)\right] \cos 2x = \left(2 A + 4 D x + 2 E\right) \cos 2x - 2 \left[(2 A x + B) + 2 (D x^2 + E x + F)\right] \sin 2x$$ For $$\sin 2x$$ term: $$\frac{d}{dx} \left[(2 D x + E) - 2 (A x^2 + B x + C)\right] \sin 2x = \left(2 D - 4 A x - 2 B\right) \sin 2x + 2 \left[(2 D x + E) - 2 (A x^2 + B x + C)\right] \cos 2x$$ Combine terms: $$y_p'' = \left[(2 A + 4 D x + 2 E) + 2 (2 D x + E) - 4 (A x^2 + B x + C)\right] \cos 2x + \left[-2 (2 A x + B) - 4 (D x^2 + E x + F) + (2 D - 4 A x - 2 B)\right] \sin 2x$$ Simplify coefficients: Cosine coefficient: $$2 A + 4 D x + 2 E + 4 D x + 2 E - 4 A x^2 - 4 B x - 4 C = -4 A x^2 + (8 D - 4 B) x + (2 A + 4 E - 4 C)$$ Sine coefficient: $$-4 A x - 2 B - 4 D x^2 - 4 E x - 4 F + 2 D - 4 A x - 2 B = -4 D x^2 + (-8 A - 4 E) x + (2 D - 4 F - 4 B)$$ 6. **Substitute into the differential equation:** $$y'' - 4 y = (x^2 - 3) \sin 2x$$ Substitute $$y_p$$ and $$y_p''$$: Left side: $$y_p'' - 4 y_p = \left[\text{cosine coeff} - 4 (A x^2 + B x + C)\right] \cos 2x + \left[\text{sine coeff} - 4 (D x^2 + E x + F)\right] \sin 2x$$ Calculate: Cosine part: $$(-4 A x^2 + (8 D - 4 B) x + (2 A + 4 E - 4 C)) - 4 (A x^2 + B x + C) = -4 A x^2 + 8 D x - 4 B x + 2 A + 4 E - 4 C - 4 A x^2 - 4 B x - 4 C$$ Simplify: $$-8 A x^2 + (8 D - 8 B) x + (2 A + 4 E - 8 C)$$ Sine part: $$(-4 D x^2 + (-8 A - 4 E) x + (2 D - 4 F - 4 B)) - 4 (D x^2 + E x + F) = -4 D x^2 - 8 A x - 4 E x + 2 D - 4 F - 4 B - 4 D x^2 - 4 E x - 4 F$$ Simplify: $$-8 D x^2 + (-8 A - 8 E) x + (2 D - 8 F - 4 B)$$ 7. **Set equal to RHS:** The RHS is: $$0 \cdot \cos 2x + (x^2 - 3) \sin 2x = 0 \cos 2x + x^2 \sin 2x - 3 \sin 2x$$ Equate coefficients: Cosine: $$-8 A x^2 + (8 D - 8 B) x + (2 A + 4 E - 8 C) = 0$$ Sine: $$-8 D x^2 + (-8 A - 8 E) x + (2 D - 8 F - 4 B) = x^2 - 3$$ 8. **Match coefficients for powers of $$x$$:** Cosine terms: - Coefficient of $$x^2$$: $$-8 A = 0 \Rightarrow A = 0$$ - Coefficient of $$x$$: $$8 D - 8 B = 0 \Rightarrow 8 D = 8 B \Rightarrow D = B$$ - Constant term: $$2 A + 4 E - 8 C = 0 \Rightarrow 0 + 4 E - 8 C = 0 \Rightarrow 4 E = 8 C \Rightarrow E = 2 C$$ Sine terms: - Coefficient of $$x^2$$: $$-8 D = 1 \Rightarrow D = -\frac{1}{8}$$ - Coefficient of $$x$$: $$-8 A - 8 E = 0 \Rightarrow 0 - 8 E = 0 \Rightarrow E = 0$$ - Constant term: $$2 D - 8 F - 4 B = -3$$ Substitute $$D = -\frac{1}{8}$$ and $$E = 0$$: $$2 \left(-\frac{1}{8}\right) - 8 F - 4 B = -3$$ $$-\frac{1}{4} - 8 F - 4 B = -3$$ $$-8 F - 4 B = -3 + \frac{1}{4} = -\frac{11}{4}$$ 9. **Use relations from cosine terms:** From cosine terms, $$D = B$$ and $$E = 2 C$$, but from sine terms $$D = -\frac{1}{8}$$ and $$E = 0$$. So, $$B = D = -\frac{1}{8}$$ $$E = 0 = 2 C \Rightarrow C = 0$$ 10. **Solve for $$F$$:** From step 8: $$-8 F - 4 B = -\frac{11}{4}$$ Substitute $$B = -\frac{1}{8}$$: $$-8 F - 4 \left(-\frac{1}{8}\right) = -\frac{11}{4}$$ $$-8 F + \frac{1}{2} = -\frac{11}{4}$$ $$-8 F = -\frac{11}{4} - \frac{1}{2} = -\frac{11}{4} - \frac{2}{4} = -\frac{13}{4}$$ $$F = \frac{13}{32}$$ 11. **Summary of coefficients:** $$A = 0, B = -\frac{1}{8}, C = 0, D = -\frac{1}{8}, E = 0, F = \frac{13}{32}$$ 12. **Write the particular solution:** $$y_p = (0 \cdot x^2 - \frac{1}{8} x + 0) \cos 2x + \left(-\frac{1}{8} x^2 + 0 \cdot x + \frac{13}{32}\right) \sin 2x$$ Simplify: $$y_p = -\frac{1}{8} x \cos 2x + \left(-\frac{1}{8} x^2 + \frac{13}{32}\right) \sin 2x$$ 13. **Final general solution:** $$y = y_c + y_p = C_1 e^{2x} + C_2 e^{-2x} - \frac{1}{8} x \cos 2x + \left(-\frac{1}{8} x^2 + \frac{13}{32}\right) \sin 2x$$