What is the solution for x^{''}+4x^'+29=0 ?

The solution for x^{''}+4x^'+29=0 is x=c_{1}+c_{2}e^{-4t}-(29t)/4

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x^{''}+4x^'+29=0

x^{^{''}} + 4 x^{^{'}} + 29 = 0

x = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 29 = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

x = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

x^{^{''}} + 3 x^{^{'}} + 2 x = cos (2 t)

y^{^{''}} + 4 y = 10 e^{5 x}

y^{^{''}} - 2 y^{^{'}} + y = sin (x) + e^{x}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = sec (t)

y^{^{''}} + y = cos (x) - sin (x)

What is the solution for x^{''}+4x^'+29=0 ?
The solution for x^{''}+4x^'+29=0 is x=c_{1}+c_{2}e^{-4t}-(29t)/4