What is the solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 ?

The solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 is x=-e^{-4t}+10e^{-4t}t

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1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 14

x = - e^{- 4 t} + 10 e^{- 4 t} t

Solution steps

\frac{1}{8} x^{''} (t) + x^{'} (t) + 2 x = 0

Solve linear ODE:

x = - e^{- 4 t} + 10 e^{- 4 t} t

x = - e^{- 4 t} + 10 e^{- 4 t} t

What is the solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 ?
The solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 is x=-e^{-4t}+10e^{-4t}t