What is the solution for y^{''}+12y^'+36y=0,y(0)=3,y^'(0)=0 ?

The solution for y^{''}+12y^'+36y=0,y(0)=3,y^'(0)=0 is y=3e^{-6t}+18te^{-6t}

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y^{''}+12y^'+36y=0,y(0)=3,y^'(0)=0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y = 3 e^{- 6 t} + 18 t e^{- 6 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{- 6 t} + 18 t e^{- 6 t}

y = 3 e^{- 6 t} + 18 t e^{- 6 t}

y^{^{''''}} - 42 y^{^{'''}} + 441 y^{^{''}} = 0

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

3 y^{^{'''}} + 17 y^{^{''}} + 31 y^{^{'}} + 14 y = 0

2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

\frac{d ^{4} y}{d x ^{4}} - 2 \frac{d ^{2} y}{d x ^{2}} + y = 0

What is the solution for y^{''}+12y^'+36y=0,y(0)=3,y^'(0)=0 ?
The solution for y^{''}+12y^'+36y=0,y(0)=3,y^'(0)=0 is y=3e^{-6t}+18te^{-6t}