What is the solution for x^{''}+2x^'+5x=8cos(2t)+2sin(2t) ?

The solution for x^{''}+2x^'+5x=8cos(2t)+2sin(2t) is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2sin(2t)

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x^{''}+2x^'+5x=8cos(2t)+2sin(2t)

x^{^{''}} + 2 x^{^{'}} + 5 x = 8 cos (2 t) + 2 sin (2 t)

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 5 x = 8 cos (2 t) + 2 sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} - y = e^{(2 x)}

y^{^{''}} + 16 y = 7 sin (x)

y^{^{'''}} - y^{^{''}} - 9 y^{^{'}} + 9 y = 7 - e^{x} + e^{3 x}

3 y^{^{''}} - y = e^{x} - 5

y^{^{''}} + 4 y^{^{'}} + 4 y = 375 \cdot t

What is the solution for x^{''}+2x^'+5x=8cos(2t)+2sin(2t) ?
The solution for x^{''}+2x^'+5x=8cos(2t)+2sin(2t) is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2sin(2t)