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))\\\\\n\nx = e^t\\\\\n\ne^{2t} y" - 2e^{t}y' + 2y = e^{2t} + \\sin(5t) \\\\\n\n\ny = f(\\ln(x)) \\\\\n\n\n\\frac{\\mathrm{d}y}{\\mathrm{d}t} = \\frac{f'(\\ln(x))}{x}\\\\\n\n\n\\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}\\\\\n\n\n\ne^{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) \\\\\n\n\ne^{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)\\\\\n\n\nf"(\\ln(x)) - f'(\\ln(x)) - 2f'(\\ln(x)) + 2f(\\ln(x)) = e^{2t} + \\sin(5t) \\\\\n\n\nf"(\\ln(x)) - 3f'(\\ln(x)) + 2f(\\ln(x)) = e^{2t} + \\sin(5t) \\\\\n\n\n\nf"(t) - 3f'(t) + 2f(t) = e^{2t} + \\sin(5t) \\\\\n\nf(t) = f_p + f_c\\\\\n\n\nf_c \\, \\textsf{is the complementary factor while}\\\\f_p \\, \\textsf{is the particular integral}\\\\\n\n\\textsf{The auxiliary equation is}\\\\\n\nm^2 - 3m + 2 = 0\\\\\n\n(m - 1)(m - 2) = 0\\\\\n\nm = 1, 2\\\\\n\n\nf"(t) - 3f'(t) + 2f(t) = e^{2t} + \\sin(5t)\\\\\n\n\n\\textsf{Solving by the method of variation of parameters}\\\\\n\nW(t) \\, \\textsf{is the Wronskian of the solution to the DE}\\\\\n\nW(t) = e^{-\\int -3 \\, \\mathrm{d}t} = e^{3t}\\\\\n\n\nV_1(t) = -\\int e^t \\frac{(e^{2t} + \\sin(5t))}{e^{3t}} \\,\\mathrm{d}t = -\\int (1 + e^{-2t}\\sin(5t))) \\,\\mathrm{d}t\\\\\n\n\n\\int e^{-2t}\\sin(5t) \\, \\mathrm{d}t = -\\frac{2e^{-2t}\\sin(5t)}{29} -\\frac{5e^{-2t}\\cos(5t)}{29}\\\\\n\n\nV_1(t) = -t + \\frac{2e^{-2t}\\sin(5t)}{29} + \\frac{5e^{-2t}\\cos(5t)}{29}\\\\\n\n\\begin{aligned}\nV_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)\\\\\n\\end{aligned}\\\\\n\n\\begin{aligned}\nf_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)\\\\\n\\end{aligned}\\\\\n\n\nf"(t) - 3f'(t) + 2f(t) = e^{2t} + \\sin(5t)\\\\\n\n\n\\therefore f(t) = Ae^t + Be^{2t} + e^{2t}(1 - t) - \\frac{15}{784}\\cos(5t) + \\frac{23}{754}\\sin(5t)\\\\\n\n\n\\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))\\\\\n\n\\textsf{Is the solution to the ODE}"