What is the solution for y^{''}+6y^'+9y=e^{-x}cos(2x) ?

The solution for y^{''}+6y^'+9y=e^{-x}cos(2x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+1/8 e^{-x}sin(2x)

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y^{''}+6y^'+9y=e^{-x}cos(2x)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{- x} cos (2 x)

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{8} e^{- x} sin (2 x)

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = e^{- x} cos (2 x)

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{8} e^{- x} sin (2 x)

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{8} e^{- x} sin (2 x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 13 y = 5 cos (2 x)

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = - 1

4 y^{^{''}} + k y = 8 sin (t)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} \cdot e^{2 t}

What is the solution for y^{''}+6y^'+9y=e^{-x}cos(2x) ?
The solution for y^{''}+6y^'+9y=e^{-x}cos(2x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+1/8 e^{-x}sin(2x)