Superposition Differential 81B713

1. **State the problem:** Solve the differential equation $$y'' - 2y - 3y = e^{2t} + 3t^2 + 4t - 5 + 5 \cos(2t)$$ using the Principle of Superposition. 2. **Rewrite the equation:** Combine like terms on the left side: $$y'' - 5y = e^{2t} + 3t^2 + 4t - 5 + 5 \cos(2t)$$ 3. **Principle of Superposition:** Since the right side is a sum of functions, the solution $$y$$ can be written as a sum of solutions to each part: $$y = y_1 + y_2 + y_3 + y_4$$ where each $$y_i$$ solves $$y'' - 5y = f_i(t)$$ with $$f_1 = e^{2t}$$, $$f_2 = 3t^2$$, $$f_3 = 4t - 5$$, and $$f_4 = 5 \cos(2t)$$. 4. **Solve the homogeneous equation:** $$y'' - 5y = 0$$ Characteristic equation: $$r^2 - 5 = 0 \implies r = \pm \sqrt{5}$$ General homogeneous solution: $$y_h = C_1 e^{\sqrt{5}t} + C_2 e^{-\sqrt{5}t}$$ 5. **Find particular solutions:** - For $$f_1 = e^{2t}$$: Try $$y_{p1} = Ae^{2t}$$. Substitute into $$y'' - 5y$$: $$4Ae^{2t} - 5Ae^{2t} = (4 - 5)Ae^{2t} = -Ae^{2t} = e^{2t}$$ So, $$-A = 1 \implies A = -1$$ Thus, $$y_{p1} = -e^{2t}$$ - For $$f_2 = 3t^2$$: Try polynomial $$y_{p2} = At^2 + Bt + C$$. Compute derivatives: $$y'_{p2} = 2At + B$$ $$y''_{p2} = 2A$$ Substitute into $$y'' - 5y$$: $$2A - 5(At^2 + Bt + C) = 3t^2$$ Equate coefficients: $$-5A = 3 \implies A = -\frac{3}{5}$$ $$-5B = 0 \implies B = 0$$ $$2A - 5C = 0 \implies 2(-\frac{3}{5}) - 5C = 0 \implies -\frac{6}{5} = 5C \implies C = -\frac{6}{25}$$ So, $$y_{p2} = -\frac{3}{5}t^2 - \frac{6}{25}$$ - For $$f_3 = 4t - 5$$: Try $$y_{p3} = Dt + E$$. Derivatives: $$y'_{p3} = D$$ $$y''_{p3} = 0$$ Substitute: $$0 - 5(Dt + E) = 4t - 5$$ Equate coefficients: $$-5D = 4 \implies D = -\frac{4}{5}$$ $$-5E = -5 \implies E = 1$$ So, $$y_{p3} = -\frac{4}{5}t + 1$$ - For $$f_4 = 5 \cos(2t)$$: Try $$y_{p4} = F \cos(2t) + G \sin(2t)$$. Derivatives: $$y'_{p4} = -2F \sin(2t) + 2G \cos(2t)$$ $$y''_{p4} = -4F \cos(2t) - 4G \sin(2t)$$ Substitute into $$y'' - 5y$$: $$(-4F \cos(2t) - 4G \sin(2t)) - 5(F \cos(2t) + G \sin(2t)) = 5 \cos(2t)$$ Simplify: $$(-4F - 5F) \cos(2t) + (-4G - 5G) \sin(2t) = 5 \cos(2t)$$ $$-9F \cos(2t) - 9G \sin(2t) = 5 \cos(2t)$$ Equate coefficients: $$-9F = 5 \implies F = -\frac{5}{9}$$ $$-9G = 0 \implies G = 0$$ So, $$y_{p4} = -\frac{5}{9} \cos(2t)$$ 6. **Combine particular solutions:** $$y_p = y_{p1} + y_{p2} + y_{p3} + y_{p4} = -e^{2t} - \frac{3}{5}t^2 - \frac{6}{25} - \frac{4}{5}t + 1 - \frac{5}{9} \cos(2t)$$ 7. **General solution:** $$y = y_h + y_p = C_1 e^{\sqrt{5}t} + C_2 e^{-\sqrt{5}t} - e^{2t} - \frac{3}{5}t^2 - \frac{6}{25} - \frac{4}{5}t + 1 - \frac{5}{9} \cos(2t)$$ This is the complete solution using the Principle of Superposition.