What is the solution for y^{''}+2y^'+5y=10cos(x) ?

The solution for y^{''}+2y^'+5y=10cos(x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+sin(x)+2cos(x)

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y^{''}+2y^'+5y=10cos(x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 10 cos (x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 10 cos (x)

Solve linear ODE:

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

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

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

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = - 6 e^{x} + 10 cos (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 6 x e^{- 3 x} + 9 + 50 sin (x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 cos (3 x)

y^{^{''}} + 4 y = 3 sin (2 t), y (0) = 2, y^{^{'}} (0) = - 1

What is the solution for y^{''}+2y^'+5y=10cos(x) ?
The solution for y^{''}+2y^'+5y=10cos(x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+sin(x)+2cos(x)