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

The solution for y^{''}-y=3x^2-1 is y=c_{1}e^x+c_{2}e^{-x}-3x^2-5

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} - 8 y = e^{2 x}

y^{^{''}} - 4 y = x e^{x} + cos (2 x)

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (5 t)

y^{^{''}} + y^{^{'}} - 2 y = 2 x, y (0) = 0, y^{^{'}} (0) = 1

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