What is the solution for x^{''}+2*x^'+x=0 ?

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

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x^{''}+2*x^'+x=0

x^{^{''}} + 2 \cdot x^{^{'}} + x = 0

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

Solution steps

x^{''} (t) + 2 \cdot x^{'} (t) + x = 0

Solve linear ODE:

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

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

0.2 x^{^{''}} + 4 x^{^{'}} + 20 x = 0

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

(D^{4} + D^{3} - 3 D^{2} - 5 D - 2) y = 0

x^{^{''}} + 5 x^{^{'}} + 6 x = 0, x (0) = 2, x^{^{'}} (0) = 3

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

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