Answer to Question #215123 in Differential Equations for ali

Question

find the relation between D and Δ of following differential equation 4(2x+1)^2 d^2y/dy^2 -4(2x+1)dy/dx + 3y =ln(2x+1))

Answer 1

"\\displaystyle\n\n4(2x+1)^2y" -4(2x+1)y' + 3y = \\ln(2x+1))\\\\\n\n\ny(x) = f(\\ln(2x + 1))\\\\\n\ny'(x) = 2\\frac{f'(\\ln(2x + 1))}{2x + 1}\\\\\n\ny''(x) = \\frac{4f''(\\ln(2x + 1))}{(2x + 1)^2} - \n\\frac{4f'(\\ln(2x + 1))}{(2x + 1)^2}\\\\\n\n16f''(\\ln(2x + 1)) - 16f'(\\ln(2x + 1)) - 8f'(\\ln(2x + 1)) + 3f(\\ln(2x + 1)) = \\ln(2x + 1)\\\\\n\n\n16f''(\\ln(2x + 1)) - 24f'(\\ln(2x + 1)) + 3f(\\ln(2x + 1)) = \\ln(2x + 1)\\\\\n\n16f''(t) - 24f'(t) + 3f(t) = t\\\\\n\n\nf(t) = c_1e^{\\frac{3 - \\sqrt{6}}{4}t} + c_2 e^{\\frac{3 + \\sqrt{6}}{4}t}\\\\\n\n\\begin{aligned}\nf(\\ln(2x + 1)) &= c_1e^{\\frac{3 - \\sqrt{6}}{4}\\ln(2x + 1)} + c_2 e^{\\frac{3 + \\sqrt{6}}{4}\\ln(2x + 1)}\n\\\\&= c_1 (2x + 1)^{\\frac{3 - \\sqrt{6}}{4}} + c_2 (2x + 1)^{\\frac{3 + \\sqrt{6}}{4}}\n\\end{aligned}\n\\\\\n\n\n16f''(t) - 24f'(t) + 3f(t) = t\\\\\n\nf(t) = At + B\\\\\n\nf'(t) = A, f"(t) = 0\\\\\n\n\\implies -24A + 3B + 3At = t\\\\\n\n\\implies 3A = 1, A = \\frac{1}{3} \\,\\,\\&\\,\\, 3B = 24A, B = \\frac{8}{3}\\\\\n\n\\implies f(t) = \\frac{t + 8}{3}, \\\\\nf(\\ln(2x + 1)) = \\frac{\\ln(2x + 1) + 8}{3}\\\\\n\n\\implies y(x) = c_1 (2x + 1)^{\\frac{3 - \\sqrt{6}}{4}} + c_2 (2x + 1)^{\\frac{3 + \\sqrt{6}}{4}} + \\frac{\\ln(2x + 1) + 8}{3}"