What is the solution for x^{''}+5x^'+4x=0 ?

The solution for x^{''}+5x^'+4x=0 is x=c_{1}e^{-t}+c_{2}e^{-4t}

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x^{''}+5x^'+4x=0

x^{^{''}} + 5 x^{^{'}} + 4 x = 0

x = c_{1} e^{- t} + c_{2} e^{- 4 t}

Solution steps

x^{''} (t) + 5 x^{'} (t) + 4 x = 0

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 4 t}

x = c_{1} e^{- t} + c_{2} e^{- 4 t}

y^{^{''}} + 4 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0, y (0) = 3, y^{^{'}} (0) = 1

y^{^{''}} - 14 y^{^{'}} + 58 y = 0

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

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for x^{''}+5x^'+4x=0 ?
The solution for x^{''}+5x^'+4x=0 is x=c_{1}e^{-t}+c_{2}e^{-4t}