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

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

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

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

y = e^{- t} + 2 e^{- t} t + 1

Solution steps

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

Solve linear ODE:

y = e^{- t} + 2 e^{- t} t + 1

y = e^{- t} + 2 e^{- t} t + 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 e^{-} 2 x - 8 sin (x) + 6 cos (x)

y^{^{''}} - 4 y = e^{3 x}

y^{^{''}} - 9 y^{^{'}} + 20 y = 3 x + e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 3 = e^{- 5}, y (0) = 0, y^{^{'}} (0) = 0

x^{^{''}} - x^{^{'}} - 2 x = 5 cos (2 t)

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