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

The solution for y^{''}+6y^'+9y=cos(x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+3/50 sin(x)+2/25 cos(x)

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

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

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{3}{50} sin (x) + \frac{2}{25} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{3}{50} sin (x) + \frac{2}{25} cos (x)

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{3}{50} sin (x) + \frac{2}{25} cos (x)

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

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 2 y = x e^{x}

(d^{2} - 1) y = x e^{x} sin (x)

y^{^{''}} + 11 y^{^{'}} + 18 y = 6 x

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

What is the solution for y^{''}+6y^'+9y=cos(x) ?
The solution for y^{''}+6y^'+9y=cos(x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+3/50 sin(x)+2/25 cos(x)