What is the solution for y^{''}+2y^'+y=1,y(0)=0,y^'(0)=0 ?

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

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y^{''}+2y^'+y=1,y(0)=0,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + y = 1, y (0) = 0, y^{^{'}} (0) = 0

y = - e^{- t} - e^{- t} t + 1

Solution steps

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

Solve linear ODE:

y = - e^{- t} - e^{- t} t + 1

y = - e^{- t} - e^{- t} t + 1

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

y^{^{''}} - 4 y^{^{'}} - 21 y = 480 e^{- t}

y^{^{''}} - 9 y^{^{'}} + 20 y = x + 1

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (2 x)

y^{^{''}} - 2 y^{^{'}} + y = e^{t}, y (0) = 0, y (0) = 5

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