What is the solution for x^{''}+2x^'+5x=1 ?

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

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x^{''}+2x^'+5x=1

x^{^{''}} + 2 x^{^{'}} + 5 x = 1

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{5}

Solution steps

x^{''} (t) + 2 x^{'} (t) + 5 x = 1

Solve linear ODE:

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{5}

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{5}

q^{^{''}} + 1100 q^{^{'}} + 100000 q = 16

(D^{2} + 2 D + 1) y = 4 sin (2 x)

x^{^{''}} - 16 x^{^{'}} + 64 x = 3 t e^{8 t}

(D^{3} + 4 D) y = e^{x} + sin (x)

y^{^{''}} + y = - cos (2 t), y^{^{'}} (0) = 0, y (0) = 0

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