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

The solution for y^{''}-y=3x^2+2x-e^x is y=c_{1}e^x+c_{2}e^{-x}-3x^2-2x-6-(e^xx)/2

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

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

y = c_{1} e^{x} + c_{2} e^{- x} - 3 x^{2} - 2 x - 6 - \frac{e ^{x} x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} - 3 x^{2} - 2 x - 6 - \frac{e ^{x} x}{2}

y = c_{1} e^{x} + c_{2} e^{- x} - 3 x^{2} - 2 x - 6 - \frac{e ^{x} x}{2}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- 1}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = - 8 x - 13, y (0) = 0, y^{^{'}} (0) = 0

2 x^{^{''}} - x^{^{'}} + 4 x = cos (3 t)

y^{^{''}} + 9 y = 5 sin (x)

y^{^{''}} + 7 y^{^{'}} + 10 y = x e^{- 2 x}

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