What is the solution for 1/8 y^{''}+y^'+2y=0,y(0)=-1,y^'(0)=12 ?

The solution for 1/8 y^{''}+y^'+2y=0,y(0)=-1,y^'(0)=12 is y=-e^{-4t}+8e^{-4t}t

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1/8 y^{''}+y^'+2y=0,y(0)=-1,y^'(0)=12

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 12

y = - e^{- 4 t} + 8 e^{- 4 t} t

Solution steps

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

Solve linear ODE:

y = - e^{- 4 t} + 8 e^{- 4 t} t

y = - e^{- 4 t} + 8 e^{- 4 t} t

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

x^{^{''}} + a x^{^{'}} = 0

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} - 20 y = 0

y^{^{''}} + 40 y^{^{'}} + 2000 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{''}} - 25 y^{^{'}} + 50 y = 0

What is the solution for 1/8 y^{''}+y^'+2y=0,y(0)=-1,y^'(0)=12 ?
The solution for 1/8 y^{''}+y^'+2y=0,y(0)=-1,y^'(0)=12 is y=-e^{-4t}+8e^{-4t}t