1. **State the problem:** Solve the differential equation $$x^2 y'' + 4x y' + 2y = e^x$$ using the method of variation of parameters. 2. **Identify the homogeneous equation:** $$x^2 y'' + 4x y' + 2y = 0$$ 3. **Solve the homogeneous equation:** Assume a solution of the form $$y = x^m$$. Substitute into the homogeneous equation: $$x^2 (m(m-1)x^{m-2}) + 4x (m x^{m-1}) + 2 x^m = 0$$ Simplify: $$m(m-1) x^m + 4m x^m + 2 x^m = 0$$ $$x^m (m^2 + 3m + 2) = 0$$ Solve the characteristic equation: $$m^2 + 3m + 2 = 0$$ $$(m+1)(m+2) = 0$$ So, $$m = -1$$ or $$m = -2$$. 4. **General solution of homogeneous equation:** $$y_h = C_1 x^{-1} + C_2 x^{-2}$$ 5. **Set up variation of parameters:** Let $$y_p = u_1(x) x^{-1} + u_2(x) x^{-2}$$ where $$u_1$$ and $$u_2$$ are functions to be determined. 6. **Formulas for variation of parameters:** The Wronskian $$W$$ of $$y_1 = x^{-1}$$ and $$y_2 = x^{-2}$$ is: $$W = y_1 y_2' - y_2 y_1' = x^{-1} (-2 x^{-3}) - x^{-2} (-1 x^{-2}) = -2 x^{-4} + x^{-4} = -x^{-4}$$ 7. **Compute $$u_1'$$ and $$u_2'$$:** $$u_1' = - \frac{y_2 g(x)}{W} = - \frac{x^{-2} e^x}{-x^{-4}} = x^{2} e^x$$ $$u_2' = \frac{y_1 g(x)}{W} = \frac{x^{-1} e^x}{-x^{-4}} = - x^{3} e^x$$ 8. **Integrate to find $$u_1$$ and $$u_2$$:** $$u_1 = \int x^{2} e^x dx$$ Use integration by parts: Let $$I = \int x^{2} e^x dx$$ First integration by parts: $$u = x^{2}, dv = e^x dx$$ $$du = 2x dx, v = e^x$$ $$I = x^{2} e^x - \int 2x e^x dx$$ Second integration by parts for $$\int 2x e^x dx$$: $$u = 2x, dv = e^x dx$$ $$du = 2 dx, v = e^x$$ $$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2 e^x + C$$ So, $$I = x^{2} e^x - (2x e^x - 2 e^x) + C = e^x (x^{2} - 2x + 2) + C$$ Therefore, $$u_1 = e^x (x^{2} - 2x + 2) + C$$ Similarly, $$u_2 = \int - x^{3} e^x dx = - \int x^{3} e^x dx$$ Use integration by parts repeatedly or recognize pattern: Let $$J = \int x^{3} e^x dx$$ First integration by parts: $$u = x^{3}, dv = e^x dx$$ $$du = 3x^{2} dx, v = e^x$$ $$J = x^{3} e^x - \int 3x^{2} e^x dx$$ We already know $$\int x^{2} e^x dx = e^x (x^{2} - 2x + 2) + C$$ So, $$J = x^{3} e^x - 3 e^x (x^{2} - 2x + 2) + C = e^x (x^{3} - 3x^{2} + 6x - 6) + C$$ Therefore, $$u_2 = - J = - e^x (x^{3} - 3x^{2} + 6x - 6) + C$$ 9. **Write the particular solution:** $$y_p = u_1 y_1 + u_2 y_2 = u_1 x^{-1} + u_2 x^{-2}$$ Substitute: $$y_p = e^x (x^{2} - 2x + 2) x^{-1} - e^x (x^{3} - 3x^{2} + 6x - 6) x^{-2}$$ Simplify: $$y_p = e^x (x - 2 + \frac{2}{x}) - e^x (x - 3 + \frac{6}{x} - \frac{6}{x^{2}})$$ $$= e^x \left[(x - 2 + \frac{2}{x}) - (x - 3 + \frac{6}{x} - \frac{6}{x^{2}})\right]$$ $$= e^x \left[x - 2 + \frac{2}{x} - x + 3 - \frac{6}{x} + \frac{6}{x^{2}}\right] = e^x \left(1 - \frac{4}{x} + \frac{6}{x^{2}}\right)$$ 10. **General solution:** $$y = y_h + y_p = C_1 x^{-1} + C_2 x^{-2} + e^x \left(1 - \frac{4}{x} + \frac{6}{x^{2}}\right)$$ --- "slug": "variation parameters", "subject": "differential equations", "desmos": {"latex": "y=C_1 x^{-1} + C_2 x^{-2} + e^x (1 - 4/x + 6/x^2)", "features": {"intercepts": true, "extrema": true}}, "q_count": 2