What is the solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-2 ?

The solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-2 is y=4e^{-t}+2te^{-t}

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

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

y = 4 e^{- t} + 2 t e^{- t}

Solution steps

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

Solve linear ODE:

y = 4 e^{- t} + 2 t e^{- t}

y = 4 e^{- t} + 2 t e^{- t}

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 9 y = 0

2 \frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - y = 0

y^{^{''}} + n y = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

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

What is the solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-2 ?
The solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-2 is y=4e^{-t}+2te^{-t}