What is the solution for y^{''}+y^'=3t^2,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+y^'=3t^2,y(0)=0,y^'(0)=1 is y=-5+5e^{-t}+t^3-3t^2+6t

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y^{''}+y^'=3t^2,y(0)=0,y^'(0)=1

y^{^{''}} + y^{^{'}} = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

y = - 5 + 5 e^{- t} + t^{3} - 3 t^{2} + 6 t

Solution steps

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

Solve linear ODE:

y = - 5 + 5 e^{- t} + t^{3} - 3 t^{2} + 6 t

y = - 5 + 5 e^{- t} + t^{3} - 3 t^{2} + 6 t

y^{^{''}} + 4 y^{^{'}} + 4 y = t - 2 e - 2 t

y^{^{''}} + y^{^{'}} = 2 sin (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 11 e^{- 3 x}

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{- 2 x}

y \cdot u^{2} - y^{^{''}} = 1

What is the solution for y^{''}+y^'=3t^2,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+y^'=3t^2,y(0)=0,y^'(0)=1 is y=-5+5e^{-t}+t^3-3t^2+6t