What is the solution for y^{''}+4y^'+3y=6x^2+x+9 ?

The solution for y^{''}+4y^'+3y=6x^2+x+9 is y=c_{1}e^{-x}+c_{2}e^{-3x}+2x^2-5x+25/3

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y^{''}+4y^'+3y=6x^2+x+9

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 x^{2} + x + 9

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + 2 x^{2} - 5 x + \frac{25}{3}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 6 x^{2} + x + 9

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + 2 x^{2} - 5 x + \frac{25}{3}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + 2 x^{2} - 5 x + \frac{25}{3}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{9 + e ^{x}}

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 θ (t) = csc^{2} (2 t)

y^{^{''}} + y = \frac{1}{cos ( x )}

y^{^{''}} + 8 y^{^{'}} = 8 x

What is the solution for y^{''}+4y^'+3y=6x^2+x+9 ?
The solution for y^{''}+4y^'+3y=6x^2+x+9 is y=c_{1}e^{-x}+c_{2}e^{-3x}+2x^2-5x+25/3