What is the solution for y^{''}+2y^'+y=0,y(0)=-3,y^'(0)=1 ?

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

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y^{''}+2y^'+y=0,y(0)=-3,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{''}} - y = 0, y^{^{'}} (0) = 0

y = \frac{d ^{2} y}{d x ^{2}}

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 8 y = 0

y^{^{''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = 0

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