What is the solution for y^{''}+8y^'+4y=0,y(0)=0 ?

The solution for y^{''}+8y^'+4y=0,y(0)=0 is y=-c_{2}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}

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y^{''}+8y^'+4y=0,y(0)=0

y^{^{''}} + 8 y^{^{'}} + 4 y = 0, y (0) = 0

y = - c_{2} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = - c_{2} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

y = - c_{2} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

16 y^{^{''}} + 56 y^{^{'}} + 49 y = 0, y (0) = 5

81 y^{^{''''}} - 625 y = 0

(d^{6} - 1) y = 0

\frac{1}{2} y^{^{''}} + 2 y^{^{'}} + 6 y = 0

y^{^{''}} + 16 y = 0, y (0) = 0, y (1) = 1

What is the solution for y^{''}+8y^'+4y=0,y(0)=0 ?
The solution for y^{''}+8y^'+4y=0,y(0)=0 is y=-c_{2}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}