What is the solution for (x^{''}+2x^'+5)^3=0 ?

The solution for (x^{''}+2x^'+5)^3=0 is x=c_{1}+c_{2}e^{-2t}-(5t)/2

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

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

x = c_{1} + c_{2} e^{- 2 t} - \frac{5 t}{2}

Solution steps

(x^{''} (t) + 2 x^{'} (t) + 5)^{3} = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- 2 t} - \frac{5 t}{2}

x = c_{1} + c_{2} e^{- 2 t} - \frac{5 t}{2}

y^{^{''}} + 9 y = 9 \cdot sec^{2} (3 t)

y^{^{''}} + 25 y = 13 sec (5 t)

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (2 x)

y^{^{''}} + 9 y^{^{'}} = tan^{3} (x)

What is the solution for (x^{''}+2x^'+5)^3=0 ?
The solution for (x^{''}+2x^'+5)^3=0 is x=c_{1}+c_{2}e^{-2t}-(5t)/2