Question

Решить линейное уравнение

y''-5y'+6y=0

Answer 1

везде одна замена:

$y = {e}^{kx}$

1)

${k}^{2} - 5k + 6 = 0 \\ d = 25 - 24 = 1 \\ k1 = \frac{5 + 1}{2} = 3 \\ k2 = 2 \\ y = c1 {e}^{2x} + c2 {e}^{3x}$

2)

${k}^{2} + 2k + 1 = 0 \\ {(k + 1)}^{2} = 0 \\ k1 = k2 = - 1 \\ y = c1 {e}^{ - x} + c2 {e}^{ - x} x$

3)

${k}^{2} + 4k + 3 = 0 \\ d = 16 - 12 = 4 \\ k1 = \frac{ - 4 + 2}{2} = - 1 \\ k2 = - 3 \\ y = c1 {e}^{ - x} + c2 {e}^{ - 3x}$