What is the solution for y^{''}+4*y^'+4*y=0 ?

The solution for y^{''}+4*y^'+4*y=0 is y=c_{1}e^{-2t}+c_{2}te^{-2t}

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

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 0

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

Solution steps

y^{''} (t) + 4 \cdot y^{'} (t) + 4 \cdot y = 0

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 1, y^{^{'}} (0) = 9

x^{^{''}} + 3 x^{^{'}} + 3 x = 0

49 y^{^{''}} - 14 y^{^{'}} + y = 0

x^{^{''}} + 2 x^{^{'}} - 3 x = 0, x (0) = 2, x^{^{'}} (0) = 3

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - 6

What is the solution for y^{''}+4*y^'+4*y=0 ?
The solution for y^{''}+4*y^'+4*y=0 is y=c_{1}e^{-2t}+c_{2}te^{-2t}