What is the solution for y^{''}+12y^'+36y=0,y(0)=2,y^'(0)=0 ?

The solution for y^{''}+12y^'+36y=0,y(0)=2,y^'(0)=0 is y=2e^{-6t}+12te^{-6t}

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y^{''}+12y^'+36y=0,y(0)=2,y^'(0)=0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = 2 e^{- 6 t} + 12 t e^{- 6 t}

Solution steps

y^{''} (t) + 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 2 e^{- 6 t} + 12 t e^{- 6 t}

y = 2 e^{- 6 t} + 12 t e^{- 6 t}

2 y^{^{''}} + 8 y = 0, y^{^{'}} (0) = 1, y (0) = 0

(D^{4} - 2 D^{3} - 3 D^{2} + 4 D + 4) y = 0

40 y^{^{''}} + 7 y^{^{'}} - 3 y = 0

8 y^{^{''}} + 8 y^{^{'}} + \frac{1}{4} y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + k^{2} \cdot y = 0

What is the solution for y^{''}+12y^'+36y=0,y(0)=2,y^'(0)=0 ?
The solution for y^{''}+12y^'+36y=0,y(0)=2,y^'(0)=0 is y=2e^{-6t}+12te^{-6t}