What is the solution for y^{''}+6y^'+9y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+6y^'+9y=0,y(0)=0,y^'(0)=1 is y=e^{-3t}t

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y^{''}+6y^'+9y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{- 3 t} t

Solution steps

y^{''} (t) + 6 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- 3 t} t

y = e^{- 3 t} t

y^{^{''}} + 4 y^{^{'}} + y 3 = 0

y^{^{''}} + 36 y = 0, y (0) = cos (6 x)

5 y^{^{''}} + 30 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 3 y = 0, y (0) = 5, y^{^{'}} (0) = 7

9 \frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 0

What is the solution for y^{''}+6y^'+9y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+6y^'+9y=0,y(0)=0,y^'(0)=1 is y=e^{-3t}t