What is the solution for y^{''}+6y^'+10y=3x-3 ?

The solution for y^{''}+6y^'+10y=3x-3 is y=e^{-3x}(c_{1}cos(x)+c_{2}sin(x))+(3x)/(10)-12/25

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y^{''}+6y^'+10y=3x-3

y^{^{''}} + 6 y^{^{'}} + 10 y = 3 x - 3

y = e^{- 3 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3 x}{10} - \frac{12}{25}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 10 y = 3 x - 3

Solve linear ODE:

y = e^{- 3 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3 x}{10} - \frac{12}{25}

y = e^{- 3 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3 x}{10} - \frac{12}{25}

x^{^{''}} + 9 x = 5 cos (t)

y^{^{''}} - 3 y^{^{'}} + 7 y = 3 e^{3 t}

y^{^{''}} + 6 y^{^{'}} = - 9

y^{^{''}} + 4 y^{^{'}} + 8 y = 24, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} + 18 y^{^{'}} - y = t^{- 1} e^{t}

What is the solution for y^{''}+6y^'+10y=3x-3 ?
The solution for y^{''}+6y^'+10y=3x-3 is y=e^{-3x}(c_{1}cos(x)+c_{2}sin(x))+(3x)/(10)-12/25