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}

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}