Answer to Question #159274 in Differential Equations for ALI HAIDER

Question

. Solve 𝑥 2 𝑑 2𝑦 𝑑2𝑥 − 2 𝑑𝑦 𝑑𝑥 − 4𝑦 = 0 

Answer 1

$\displaystyle x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\frac{\mathrm{d}y}{\mathrm{d}x} - 4y = 0\\ \textsf{Let}\,\, y = f(\ln{x})\\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\ln{x})}{x}\\ \begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= -\frac{f'(\ln{x})}{x^2} + \frac{f''(\ln{x})}{x^2} &= \frac{1}{x^2}\left(f''(\ln{x} - f'(\ln{x})\right) \end{aligned} \\ x^2\left(\frac{1}{x^2}\left(f''(\ln{x}) - f'(\ln{x})\right)\right) - 2x\cdot\frac{f'(\ln{x})}{x} - 4f(\ln{x}) = 0\\ f''(\ln{x}) - f'(\ln{x} - 2f'(\ln{x}) - 4f(\ln{x}) = 0\\ f''(\ln{x}) - 3f'(\ln{x}) - 4f(\ln{x}) = 0\\ \textsf{Let}\,\, t(l) = f(\ln{x})\\ t'' - 3t' - 4t = 0\\ \textsf{Auxiliary equation}; \,\,m^2 - 3m - 4 = 0\\ m = -1, 4\\ t = Ae^{-l} + Be^{4l} \\ t = Ae^{-\ln{x}} + Be^{4\ln{x}} \\ t(l) = f(\ln{x}) = y = \frac{A}{x} + Bx^4 \\ \therefore y = \frac{A}{x} + Bx^4$