Решить ДУ: y''-y=0 и найти значение y, если y(0)=1 и y'(0)=-1

Y''-y=0 =>
λ²-1=0 => λ²=1 => λ₁,₂ = +/- 1 .
$y=C₁e^{λ₁x}+C₂e^{λ₂x}$ =>
$y = C₁e^{x}+ C₂e^{-x}$ =>
$y' = C₁e^{x}-C₂e^{-x}$ => при y(0) = 1 и y'(0)=-1 =>
y(0)=C₁+C₂=1 b y'(0)=C₁-C₂=-1 =>
$\left \{ {{C₁+C₂=1} \atop {C₁-C₂=-1}} \right.$ =>
C₁ = 0 и C₂ = -1. => $y=e^{-x}$