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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} - 5 x = e^{- t}

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

y^{^{''}} - a y = - c

y^{^{''}} + 36 y = 60 sec (6 t)

2 x^{^{''}} + 4 x^{^{'}} - 3 = 0

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