What is the solution for y^{''}+y=t^3 ?

The solution for y^{''}+y=t^3 is y=c_{1}cos(t)+c_{2}sin(t)+t^3-6t

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y^{''}+y=t^3

y^{^{''}} + y = t^{3}

y = c_{1} cos (t) + c_{2} sin (t) + t^{3} - 6 t

Solution steps

y^{''} (t) + y = t^{3}

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + t^{3} - 6 t

y = c_{1} cos (t) + c_{2} sin (t) + t^{3} - 6 t

y^{^{''}} - 8 y^{^{'}} + 16 y = x

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 t^{2} e^{5 t}

y^{^{''}} - 12 y^{^{'}} + 36 y = 18 x + 2

y^{^{''}} - 5 y = 2 x^{3} - 4 x^{2} - x + 6

What is the solution for y^{''}+y=t^3 ?
The solution for y^{''}+y=t^3 is y=c_{1}cos(t)+c_{2}sin(t)+t^3-6t