Question

y"+3y'+2y=0
н.у y(0)=-1 y'(0)=3

Answer 1

Ответ:

y = C₁ e^(-2x) + C₂ e^(-x)  

y(0) = C₁ + C₂, C₁ + C₂ = - 1, C₁ = - 1 - C₂

y` = - 2*C₁e^(-2x) - C₂e^(-x) , y`(0) = 3

- 2*C₁ - C₂ = 3,  

- 2*(-1 - C₂) - C₂ = 3

2 + 2C₂ - C₂ = 3

C₂ = 1

C₁ = - 1 - 1 = - 2

y = e^(-2x) - 2e^(-x)  

y = 1/(e^2x) - 2/(e^x)

надеюсь так...