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

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