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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 2

y^{^{'''}} + 19 y^{^{''}} - 14 y^{^{'}} - 266 y = 0

40 y^{^{''}} - a y^{^{'}} + 4000 y = 0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 28 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 0

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