Answer to Question #204569 in Differential Equations for Usman Ali

Question

(𝑡 +2)^2 𝑦′′ + (𝑡+2) 𝑦′ + 𝑦=0

Answer 1

$\displaystyle (t + 2)^2 y'' + (t + 2)y' + y = 0\\ y = f(\ln(t + 2)) = f(x)\\ y' = \frac{f'(\ln(t + 2))}{t + 2} \\ y'' = \frac{f''(\ln(t + 2))}{(t + 2)^2} - \frac{f'(\ln(t + 2))}{(t + 2)^2}\\ f''(\ln(t + 2)) - f'(\ln(t + 2)) + f'(\ln(t + 2)) + f(\ln(t + 2)) = 0\\ f''(\ln(t + 2)) + f(\ln(t + 2)) = 0\\ f''(x) + f(x) = 0\\ f(x) = c_1\cos{x} + c_2\sin{x}\\ f(x) = c_1\cos(\ln(t + 2)) + c_2\sin(\ln(t + 2))$