Question

y-6y'+5y=0 y(0)=1 y'(0)=-1

Answer 1

$y(x) = {e}^{mx} \\y'(x) = m {e}^{mx} \\ y''(x) = {m}^{2} {e}^{mx}$

${m}^{2} - 5m + 6 = 0 \\ Δ =25 - 24 = 1 \\ m = \frac{5 ±1}{2}$

$m' = \frac{5 + 1}{2} = \frac{6}{2} = 3 \\ m'' = \frac{5 - 1}{2} = \frac{4}{2} = 2$

$y(x) = c1 {e}^{3x} + c2 {e}^{2x} \\y'(x) = 3c1 {e}^{3x} + 2c2 {e}^{2x}$

$y(0) = c1 {e}^{3.0} + c2 {e}^{2.0} \\ 1 = c1 + c2 \\ c1 + c2 = 1$

$y'(0) = 3c1 {e}^{3.0} + 2c2 {e}^{2.0} \\ 3c1 + 2c2 = - 1$

{c1+c2=1×(-2)

{3c1+2c2=-1

{–2c1–2c2=–2

+ {3c1+2c2=-1

c1=-3

c1+c2=1

–3+c2=1

c2=1+3

c2=4

$\frac{c1}{c2} = \frac{ - 3}{4}$

Nenhuma das alternativas pede revisão de questão