Question

1) y^11-9y^1=0
2) y^11-2y^1+y=0
Помогите пожалуйста

Answer 1

Ответ:

Пошаговое объяснение:

$y'' - 9y' = 0\y'' = 9y'\y' = u \u' = 9u \frac{du}{dx} = 9u\frac{du}{u} = 9dx\ln(u) = 9x + C_{1} \y' = 9x + C_{1}\frac{dy}{dx} = 9x + C_{1}\dy = (9x + C_{1})dx\y = frac{9x^2}{2} +C_{1}x + C_{2}$

$y'' -2y' + 1 = 0\a^2 - 2a + 1 =0\(a-1)^2 = 0\a_{1} = 1, k_a_{1} =2\a>0, k_a_{1} =2 => y(x) = C_1e^{1x}+C_2xe^{1x}$