What is the solution for x^{''}+4x^'+29=0,x(0)=5,x^'(0)=5 ?

The solution for x^{''}+4x^'+29=0,x(0)=5,x^'(0)=5 is x=5+49/16-49/16 e^{-4t}-(29t)/4

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x^{''}+4x^'+29=0,x(0)=5,x^'(0)=5

x^{^{''}} + 4 x^{^{'}} + 29 = 0, x (0) = 5, x^{^{'}} (0) = 5

x = 5 + \frac{49}{16} - \frac{49}{16} e^{- 4 t} - \frac{29 t}{4}

Solution steps

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

Solve linear ODE:

x = 5 + \frac{49}{16} - \frac{49}{16} e^{- 4 t} - \frac{29 t}{4}

x = 5 + \frac{49}{16} - \frac{49}{16} e^{- 4 t} - \frac{29 t}{4}

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 e^{- 2 t}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 1

y^{^{''}} - y = e^{a t}

2 y^{^{''}} - 6 y^{^{'}} = 5

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

What is the solution for x^{''}+4x^'+29=0,x(0)=5,x^'(0)=5 ?
The solution for x^{''}+4x^'+29=0,x(0)=5,x^'(0)=5 is x=5+49/16-49/16 e^{-4t}-(29t)/4