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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 6 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = t e^{t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = x - 6

x^{^{''}} + x = cos (3 t)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 5 + 4 sin (x)

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