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

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

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

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

x = 2 e^{- t} + 3 t e^{- t}

Solution steps

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

Solve linear ODE:

x = 2 e^{- t} + 3 t e^{- t}

x = 2 e^{- t} + 3 t e^{- t}

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