Diff Eq Cosine 6De5C6

1. **Problem:** Solve the differential equation $$\frac{d^2 x}{dt^2} - 2 \frac{dx}{dt} + 5x = 34 \cos 2t$$ with initial conditions $$x(0) = 10$$ and $$\frac{dx}{dt}(0) = 2$$. 2. **Step 1: Solve the homogeneous equation** The associated homogeneous equation is: $$\frac{d^2 x}{dt^2} - 2 \frac{dx}{dt} + 5x = 0$$ The characteristic equation is: $$r^2 - 2r + 5 = 0$$ Using the quadratic formula: $$r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i$$ 3. **Step 2: Write the homogeneous solution** Since roots are complex conjugates $$1 \pm 2i$$, the homogeneous solution is: $$x_h = e^{t}(C_1 \cos 2t + C_2 \sin 2t)$$ 4. **Step 3: Find a particular solution** The right side is $$34 \cos 2t$$, which suggests trying a particular solution of the form: $$x_p = A \cos 2t + B \sin 2t$$ 5. **Step 4: Compute derivatives of $$x_p$$** $$\frac{dx_p}{dt} = -2A \sin 2t + 2B \cos 2t$$ $$\frac{d^2 x_p}{dt^2} = -4A \cos 2t - 4B \sin 2t$$ 6. **Step 5: Substitute into the differential equation** Substitute $$x_p$$ and its derivatives: $$(-4A \cos 2t - 4B \sin 2t) - 2(-2A \sin 2t + 2B \cos 2t) + 5(A \cos 2t + B \sin 2t) = 34 \cos 2t$$ Simplify: $$(-4A \cos 2t - 4B \sin 2t) + 4A \sin 2t - 4B \cos 2t + 5A \cos 2t + 5B \sin 2t = 34 \cos 2t$$ Group terms: $$(-4A - 4B + 5A) \cos 2t + (-4B + 4A + 5B) \sin 2t = 34 \cos 2t$$ Simplify coefficients: $$ (A - 4B) \cos 2t + (4A + B) \sin 2t = 34 \cos 2t$$ 7. **Step 6: Equate coefficients** From $$\cos 2t$$: $$A - 4B = 34$$ From $$\sin 2t$$: $$4A + B = 0$$ 8. **Step 7: Solve the system** From second equation: $$B = -4A$$ Substitute into first: $$A - 4(-4A) = 34 \Rightarrow A + 16A = 34 \Rightarrow 17A = 34 \Rightarrow A = 2$$ Then: $$B = -4 \times 2 = -8$$ 9. **Step 8: Write the general solution** $$x = x_h + x_p = e^{t}(C_1 \cos 2t + C_2 \sin 2t) + 2 \cos 2t - 8 \sin 2t$$ 10. **Step 9: Apply initial conditions** At $$t=0$$: $$x(0) = e^{0}(C_1 \cdot 1 + C_2 \cdot 0) + 2 \cdot 1 - 8 \cdot 0 = C_1 + 2 = 10 \Rightarrow C_1 = 8$$ 11. **Step 10: Compute $$\frac{dx}{dt}$$** $$\frac{dx}{dt} = \frac{d}{dt} \left[ e^{t}(C_1 \cos 2t + C_2 \sin 2t) + 2 \cos 2t - 8 \sin 2t \right]$$ Use product rule: $$\frac{d}{dt} e^{t}(C_1 \cos 2t + C_2 \sin 2t) = e^{t}(C_1 \cos 2t + C_2 \sin 2t) + e^{t}(-2 C_1 \sin 2t + 2 C_2 \cos 2t)$$ Derivatives of particular solution: $$\frac{d}{dt} (2 \cos 2t - 8 \sin 2t) = -4 \sin 2t - 16 \cos 2t$$ So: $$\frac{dx}{dt} = e^{t}(C_1 \cos 2t + C_2 \sin 2t) + e^{t}(-2 C_1 \sin 2t + 2 C_2 \cos 2t) - 4 \sin 2t - 16 \cos 2t$$ 12. **Step 11: Evaluate at $$t=0$$** $$\frac{dx}{dt}(0) = e^{0}(C_1 \cdot 1 + C_2 \cdot 0) + e^{0}(-2 C_1 \cdot 0 + 2 C_2 \cdot 1) - 4 \cdot 0 - 16 \cdot 1 = C_1 + 2 C_2 - 16 = 2$$ Substitute $$C_1 = 8$$: $$8 + 2 C_2 - 16 = 2 \Rightarrow 2 C_2 - 8 = 2 \Rightarrow 2 C_2 = 10 \Rightarrow C_2 = 5$$ 13. **Final solution:** $$\boxed{x = e^{t}(8 \cos 2t + 5 \sin 2t) + 2 \cos 2t - 8 \sin 2t}$$ This completes the solution for the first problem.