Question

A second order homogenous differential equation with constant coefficient and polynomial
coefficient is given, apply Euler-Cauchy equation to solve the given differential equation by
using the provided initial value problem (IVP).
x^2 y'' +8y' +10y =0 y(1) = 0 , y'(1) = 2.5

Answer 1

$\displaystyle x^2 y'' +8xy' +10y =0,\,\, y(1) = 0,\,\, y'(1) = 2.5\\ \textsf{Let}\,\,y = f(\ln{x}), y' = \frac{f'(\ln{x})}{x}\\ y'' = \frac{f''(\ln{x})}{x^2} - \frac{f'(\ln{x})}{x^2}\\ f"(\ln{x}) - f'(\ln{x}) + \frac{8f'(\ln{x})}{x}\times x + 10f(\ln{x}) = 0\\ f"(\ln{x}) - f'(\ln{x}) + 8f'(\ln{x}) + 10f(\ln{x}) = 0\\ f"(\ln{x}) + 7f'(\ln{x}) + 10f(\ln{x}) = 0\\ \textsf{Let}\,\, t = \ln{x}\\ f"(t) + 7f'(t) + 10f(t) = 0\\ \textsf{The auxiliary equation is}\\ m^2 + 7m + 10 = 0\\ (m + 5)(m + 2) = 0\\ m = -2, -5\\ f(t) = Ae^{-2t} + Be^{-5t}\,\, \textsf{is the general solution}\\ f(\ln{x}) = Ae^{-2\ln{x}} + Be^{-5\ln{x}}\\ y(x) = f(\ln{x}) = \frac{A}{x^2} + \frac{B}{x^5}\\ y(1) = A + B = 0\\ y'(x) = -2\frac{A}{x^3} - 5\frac{B}{x^6}\\ y'(1) = -2A - 5B = \frac{5}{2}\\ \textsf{But}\,\, A = -B\\ -2A + 5A = \frac{5}{2}\\ 3A = \frac{5}{2}\\ A = \frac{5}{6}, B = -\frac{5}{6}\\ y(x) = \frac{5}{6x^2} - \frac{5}{6x^5}$