What is the solution for x^{''}+3x^'+2x=0,x^'(0)=2 ?

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

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

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

x = (- 2 - 2 c_{2}) e^{- t} + c_{2} e^{- 2 t}

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = 0

Solve linear ODE:

x = (- 2 - 2 c_{2}) e^{- t} + c_{2} e^{- 2 t}

x = (- 2 - 2 c_{2}) e^{- t} + c_{2} e^{- 2 t}

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