What is the solution for y^{''}+2y^'+5y=cos(2x)-4sin(2x) ?

The solution for y^{''}+2y^'+5y=cos(2x)-4sin(2x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+cos(2x)

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y^{''}+2y^'+5y=cos(2x)-4sin(2x)

y^{^{''}} + 2 y^{^{'}} + 5 y = cos (2 x) - 4 sin (2 x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = cos (2 x) - 4 sin (2 x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = t^{5} e^{- 5 t}

y^{^{''}} + 16 y = 9 (t) - 9 (t - 98 π), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{- x} + 9

y^{^{''}} + 18 y^{^{'}} + 82 y = 2 e^{- 9 x} cos (x)

y^{^{''}} + 9 y^{^{'}} = t^{2} e^{3 t} + 6

What is the solution for y^{''}+2y^'+5y=cos(2x)-4sin(2x) ?
The solution for y^{''}+2y^'+5y=cos(2x)-4sin(2x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+cos(2x)