What is the solution for x^{''}+6x^'+8x=0,x(0)=1,x^'(0)=2 ?

The solution for x^{''}+6x^'+8x=0,x(0)=1,x^'(0)=2 is x=3e^{-2t}-2e^{-4t}

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x^{''}+6x^'+8x=0,x(0)=1,x^'(0)=2

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 1, x^{^{'}} (0) = 2

x = 3 e^{- 2 t} - 2 e^{- 4 t}

Solution steps

x^{''} (t) + 6 x^{'} (t) + 8 x = 0

Solve linear ODE:

x = 3 e^{- 2 t} - 2 e^{- 4 t}

x = 3 e^{- 2 t} - 2 e^{- 4 t}

What is the solution for x^{''}+6x^'+8x=0,x(0)=1,x^'(0)=2 ?
The solution for x^{''}+6x^'+8x=0,x(0)=1,x^'(0)=2 is x=3e^{-2t}-2e^{-4t}