What is the solution for y^{''}+3y=6t^3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+3y=6t^3,y(0)=0,y^'(0)=0 is y=2t^3-4t+4/(sqrt(3))sin(sqrt(3)t)

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y^{''}+3y=6t^3,y(0)=0,y^'(0)=0

y^{^{''}} + 3 y = 6 t^{3}, y (0) = 0, y^{^{'}} (0) = 0

y = 2 t^{3} - 4 t + \frac{4}{3} sin (3 t)

Solution steps

y^{''} (t) + 3 y = 6 t^{3}

Solve linear ODE:

y = 2 t^{3} - 4 t + \frac{4}{3} sin (3 t)

y = 2 t^{3} - 4 t + \frac{4}{3} sin (3 t)

y^{^{''}} - 10 y^{^{'}} + 41 y = 68 e^{4 t}

y^{^{''}} + 25 y = 8 cos (5 x)

y^{^{''}} - 7 y^{^{'}} + 10 y = 8 e^{x}

y^{^{''}} + a^{2} y = b cos (c x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 2 y = \frac{1}{2} e^{2 t} + \frac{1}{2} e^{- 2 t}

What is the solution for y^{''}+3y=6t^3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y=6t^3,y(0)=0,y^'(0)=0 is y=2t^3-4t+4/(sqrt(3))sin(sqrt(3)t)