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

The solution for y^{''}+6y^'+9y=2sin(x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+4/25 sin(x)-3/25 cos(x)

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y^{''}+6y^'+9y=2sin(x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 sin (x)

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{4}{25} sin (x) - \frac{3}{25} cos (x)

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = 2 sin (x)

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{4}{25} sin (x) - \frac{3}{25} cos (x)

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{4}{25} sin (x) - \frac{3}{25} cos (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x ^{2} + 1}

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} cos (4 x)

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

y^{^{''}} - 4 y^{^{'}} - 12 y = 12 t, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}+6y^'+9y=2sin(x) ?
The solution for y^{''}+6y^'+9y=2sin(x) is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+4/25 sin(x)-3/25 cos(x)