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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 3 D + 2) y = 0

y^{^{''}} + 16 π^{2} y = 0

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

t^{''} + 4 t (x (t, z)) = 0, t (0) = - 5, t^{^{'}} (0) = 2

x^{^{''}} + 2 x^{^{'}} + 26 x = 0, x (0) = 1, x^{^{'}} (0) = 4

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