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

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

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

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

y = c_{1} + c_{2} e^{- x} + \frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 7 x

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 7 x

y = c_{1} + c_{2} e^{- x} + \frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 7 x

y^{^{''}} + y = 2 + t

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 8 sin (x) - 4 cos (x)

y^{^{''}} + 9 y = sin^{4} (x)

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 4 x = 7 - 2 t^{2}

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