Question

9y''+6y'+y=0
y(0)=-5;y'(0)=4

Answer 1

9y''+6y'+y=0
y(0)=-5;y'(0)=4

9k²+6k+1=0

(3k+1)²=0 ⇔ k1=k2=-1/3  

У₁=e^(-x/3)    У₂=x·e^(-x/3)

Y₀₀=C₁Y₁+C₂Y₂   Y₀₀=C₁e^(-x/3)+C₂x·e^(-x/3)=e^(-x/3)(C₁+C₂x)

Y'₀₀=[e^(-x/3)]'(C₁+C₂x)+[e^(-x/3)](C₁+C₂x)'=
= -([e^(-x/3)]/3 ·(C₁+C₂x)+[e^(-x/3)]C₂

y(0)=-5;y'(0)=4

Y₀₀(0)=C₁= -5                                C₁= -5   
Y'₀₀(0)=-(1/3) ·(C₁)+C₂=4              C₂=4+(-5/3)=7/3

Y₀₀=-5e^(-x/3)+(7/3)x·e^(-x/3)=e^(-x/3)(-5+7x/3)