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

The solution for y^{''}+2y^'+y=t^2,y(0)=1,y^'(0)=1 is y=t^2-4t-5e^{-t}+6

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

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

y = t^{2} - 4 t - 5 e^{- t} + 6

Solution steps

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

Solve linear ODE:

y = t^{2} - 4 t - 5 e^{- t} + 6

y = t^{2} - 4 t - 5 e^{- t} + 6

y^{^{''}} + 4 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 6

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 5

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 x + 4

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 e^{- 2 t}

What is the solution for y^{''}+2y^'+y=t^2,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+2y^'+y=t^2,y(0)=1,y^'(0)=1 is y=t^2-4t-5e^{-t}+6