What is the solution for y^{''}+2y^'+5y=8cos(x) ?

The solution for y^{''}+2y^'+5y=8cos(x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+4/5 sin(x)+8/5 cos(x)

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y^{''}+2y^'+5y=8cos(x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 8 cos (x)

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{4}{5} sin (x) + \frac{8}{5} cos (x)

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 8 cos (x)

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{4}{5} sin (x) + \frac{8}{5} cos (x)

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{4}{5} sin (x) + \frac{8}{5} cos (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = (1 - x) e^{4 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 1 + x e^{x}

4 y^{^{''}} + y = cos (x)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 5

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

What is the solution for y^{''}+2y^'+5y=8cos(x) ?
The solution for y^{''}+2y^'+5y=8cos(x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+4/5 sin(x)+8/5 cos(x)