What is the solution for x^{''}+2x^'+5x=5cos(t) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 4 csc^{2} (2 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = sin (2 t)

y^{^{''}} + 4 y^{^{'}} + 2 y = x e^{- 2 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 5 t} ln (t)

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

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