What is the solution for y^{''}+2y^'+y=0,y(0)=2,y^'(0)=10 ?

The solution for y^{''}+2y^'+y=0,y(0)=2,y^'(0)=10 is y=2e^{-t}+12te^{-t}

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y^{''}+2y^'+y=0,y(0)=2,y^'(0)=10

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 10

y = 2 e^{- t} + 12 t e^{- t}

Solution steps

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

Solve linear ODE:

y = 2 e^{- t} + 12 t e^{- t}

y = 2 e^{- t} + 12 t e^{- t}

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} = 7 y^{^{'}} - 12 y

(D^{2} + 4) y = 0

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

y^{^{''''}} - 8 y^{^{'''}} = 0

What is the solution for y^{''}+2y^'+y=0,y(0)=2,y^'(0)=10 ?
The solution for y^{''}+2y^'+y=0,y(0)=2,y^'(0)=10 is y=2e^{-t}+12te^{-t}