What is the solution for y^{''}-y^'-2y=-2x^3-3x^2+8x+1 ?

The solution for y^{''}-y^'-2y=-2x^3-3x^2+8x+1 is y=c_{1}e^{2x}+c_{2}e^{-x}+x^3-x

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y^{''}-y^'-2y=-2x^3-3x^2+8x+1

y^{^{''}} - y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

Solve linear ODE:

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

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

y^{^{''}} - 9 y = \frac{9 x}{e ^{3 x}}

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

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

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

y^{^{''}} - y^{^{'}} = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-y^'-2y=-2x^3-3x^2+8x+1 ?
The solution for y^{''}-y^'-2y=-2x^3-3x^2+8x+1 is y=c_{1}e^{2x}+c_{2}e^{-x}+x^3-x