What is the solution for (d^2x)/(dt^2)+2(dx)/(dt)+5x=1 ?

The solution for (d^2x)/(dt^2)+2(dx)/(dt)+5x=1 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+1/5

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(d^2x)/(dt^2)+2(dx)/(dt)+5x=1

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 5 x = 1

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

Solution steps

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

y^{^{''}} + y^{^{'}} = c

(D^{3} - 5 D^{2} + 8 D - 4) y = e^{2 x} + 2 e^{x} + 3 e^{- x}

y^{^{''}} - 16 y = 16 t - 32 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 51

y^{^{''}} + 2 y^{^{'}} + 1000 y = 2 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 cos (x)

What is the solution for (d^2x)/(dt^2)+2(dx)/(dt)+5x=1 ?
The solution for (d^2x)/(dt^2)+2(dx)/(dt)+5x=1 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+1/5