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

The solution for y^{''}+3y^'+2y=4x^2 is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x^2-6x+7

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x^{2} - 6 x + 7

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x^{2} - 6 x + 7

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x^{2} - 6 x + 7

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{- 2 t}

(D^{2} - 4 D + 13) y = 24 e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} + 10 y = 3 e^{x} tan (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 7 e^{- 2 t}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8 e^{- t}

What is the solution for y^{''}+3y^'+2y=4x^2 ?
The solution for y^{''}+3y^'+2y=4x^2 is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x^2-6x+7