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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 5

u^{^{''}} = H (t)

0.25 \cdot 0.04 x^{^{''}} + 2 \cdot 0.04 x^{^{'}} + x = 0

y^{^{''}} + 20 y^{^{'}} + 92 y = 0

x^{^{''}} + 3 x^{^{'}} + 2 x = 0, x (0) = 0, x^{^{'}} (0) = 0

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