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 . [latex]y=C₁e^{λ₁x}+C₂e^{λ₂x}[/latex] => [latex]y = C₁e^{x}+ C₂e^{-x}[/latex] => [latex]y' = C₁e^{x}-C₂e^{-x} [/latex] => при y(0) = 1 и y'(0)=-1 => y(0)=C₁+C₂=1 b y'(0)=C₁-C₂=-1 => [latex] \left \{ {{C₁+C₂=1} \atop {C₁-C₂=-1}} \right. [/latex] => C₁ = 0 и C₂ = -1. => [latex]y=e^{-x} [/latex]