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

The solution for y^{''}+3y=x^3-1 is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)+(x^3)/3-(2x)/3-1/3

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x ^{3}}{3} - \frac{2 x}{3} - \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x ^{3}}{3} - \frac{2 x}{3} - \frac{1}{3}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{x ^{3}}{3} - \frac{2 x}{3} - \frac{1}{3}

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} = x, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2}

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

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

What is the solution for y^{''}+3y=x^3-1 ?
The solution for y^{''}+3y=x^3-1 is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)+(x^3)/3-(2x)/3-1/3