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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 3 sin (x) + cos (3 x)

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 - 2 x

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 54 x + 18

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 5 e^{2 x}

y^{^{''}} - y^{^{'}} - 2 y = 7 x + 4

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