What is the solution for y^{''}+3y=3t^3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+3y=3t^3,y(0)=0,y^'(0)=0 is y=t^3-2t+2/(sqrt(3))sin(sqrt(3)t)

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y^{''}+3y=3t^3,y(0)=0,y^'(0)=0

y^{^{''}} + 3 y = 3 t^{3}, y (0) = 0, y^{^{'}} (0) = 0

y = t^{3} - 2 t + \frac{2}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

y = t^{3} - 2 t + \frac{2}{3} sin (3 t)

y = t^{3} - 2 t + \frac{2}{3} sin (3 t)

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

y^{^{''}} - 16 y = sinh (4 x)

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

x^{^{''}} + 4 x = sin^{2} (2 t)

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

What is the solution for y^{''}+3y=3t^3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y=3t^3,y(0)=0,y^'(0)=0 is y=t^3-2t+2/(sqrt(3))sin(sqrt(3)t)