Answer to Question #139455 in Differential Equations for Nikhil

Question

Solve
x^2y"- 2xy' + 2y=x^2+sin(5Inx)

Answer 1

$\displaystyle x^2y"- 2xy' + 2y=x^2+\sin(\ln(5x))\\ x = e^t\\ e^{2t} y" - 2e^{t}y' + 2y = e^{2t} + \sin(5t) \\ y = f(\ln(x)) \\ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{f'(\ln(x))}{x}\\ \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = \frac{-f'(\ln(x))}{x^2} + \frac{f''(\ln(x))}{x^2} = \frac{f"(\ln(x)) - f'(\ln(x))}{x^2}\\ e^{2t}\left(\frac{f"(\ln(x)) - f'(\ln(x))}{x^2}\right) - 2e^{t}\frac{f'(\ln(x))}{x} + 2f(\ln(x)) = e^{2t} + \sin(5t) \\ e^{2t}\left(\frac{f"(\ln(x)) - f'(\ln(x))}{e^{2t}}\right) - 2e^{t}\frac{f'(\ln(x))}{e^t} + 2f(\ln(x)) = e^{2t} + \sin(5t)\\ f"(\ln(x)) - f'(\ln(x)) - 2f'(\ln(x)) + 2f(\ln(x)) = e^{2t} + \sin(5t) \\ f"(\ln(x)) - 3f'(\ln(x)) + 2f(\ln(x)) = e^{2t} + \sin(5t) \\ f"(t) - 3f'(t) + 2f(t) = e^{2t} + \sin(5t) \\ f(t) = f_p + f_c\\ f_c \, \textsf{is the complementary factor while}\\f_p \, \textsf{is the particular integral}\\ \textsf{The auxiliary equation is}\\ m^2 - 3m + 2 = 0\\ (m - 1)(m - 2) = 0\\ m = 1, 2\\ f"(t) - 3f'(t) + 2f(t) = e^{2t} + \sin(5t)\\ \textsf{Solving by the method of variation of parameters}\\ W(t) \, \textsf{is the Wronskian of the solution to the DE}\\ W(t) = e^{-\int -3 \, \mathrm{d}t} = e^{3t}\\ V_1(t) = -\int e^t \frac{(e^{2t} + \sin(5t))}{e^{3t}} \,\mathrm{d}t = -\int (1 + e^{-2t}\sin(5t))) \,\mathrm{d}t\\ \int e^{-2t}\sin(5t) \, \mathrm{d}t = -\frac{2e^{-2t}\sin(5t)}{29} -\frac{5e^{-2t}\cos(5t)}{29}\\ V_1(t) = -t + \frac{2e^{-2t}\sin(5t)}{29} + \frac{5e^{-2t}\cos(5t)}{29}\\ \begin{aligned} V_2(t) &= \int e^{2t} \frac{(e^{2t} + \sin(5t))}{e^{3t}} \,\mathrm{d}t \\&= \int e^t + e^{-t}\sin(5t) \,\mathrm{d}t \\&= e^t - \frac{5}{26}e^{-t} \cos(5t) - \frac{1}{26} e^{-t}\sin(5t)\\ \end{aligned}\\ \begin{aligned} f_p &= \left(-t + \frac{2e^{-2t}\sin(5t)}{29} + \frac{5e^{-2t}\cos(5t)}{29}\right)e^{2t} \\&+ \left(e^t - \frac{5}{26}e^{-t} \cos(5t) - \frac{1}{26} e^{-t}\sin(5t)\right)e^t \\&= -te^{2t}+ \frac{2\sin(5t)}{29} + \frac{5\cos(5t)}{29} + e^{2t} - \frac{5}{26} \cos(5t) - \frac{1}{26}\sin(5t) \\&= e^{2t}(1 - t) - \frac{15}{784}\cos(5t) + \frac{23}{754}\sin(5t)\\ \end{aligned}\\ f"(t) - 3f'(t) + 2f(t) = e^{2t} + \sin(5t)\\ \therefore f(t) = Ae^t + Be^{2t} + e^{2t}(1 - t) - \frac{15}{784}\cos(5t) + \frac{23}{754}\sin(5t)\\ \therefore y = f(x) = Ax + Bx^2 + x^2(1 - \ln(x)) - \frac{15}{784}\cos(5\ln(x)) + \frac{23}{754}\sin(5\ln(x))\\ \textsf{Is the solution to the ODE}$