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

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

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

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

x = e^{- t} t

Solution steps

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

Solve linear ODE:

x = e^{- t} t

x = e^{- t} t

x^{^{''}} + 28 x^{^{'}} + 196 x = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 0, y (0) = 1, y (1) = 0

y^{^{''}} + 3 y^{^{'}} = - 2 y

y^{^{'''}} - 4 y^{^{''}} - 32 y^{^{'}} = 0

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

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