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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

(D^{2} + 1) y = sec^{2} (x)

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

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

2 y^{^{''}} - y^{^{'}} - 3 y = 3 x e^{2 x}

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