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

The solution for y^{''}+2y^'-8y=0,y(0)=-2,y^'(0)=1 is y=(-2+5/6)e^{2t}-5/6 e^{-4t}

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

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

y = (- 2 + \frac{5}{6}) e^{2 t} - \frac{5}{6} e^{- 4 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = (- 2 + \frac{5}{6}) e^{2 t} - \frac{5}{6} e^{- 4 t}

y = (- 2 + \frac{5}{6}) e^{2 t} - \frac{5}{6} e^{- 4 t}

y^{^{''}} + 0 y^{^{'}} + 16 y = 0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

y^{^{'''}} - 2 y^{^{''}} - 63 y^{^{'}} = 0

\frac{3}{4} x^{^{''}} + 72 x = 0, x (0) = - \frac{1}{4}, x^{^{'}} (0) = 0

\frac{1}{16} x^{^{''}} + x^{^{'}} + 4 x = 0

What is the solution for y^{''}+2y^'-8y=0,y(0)=-2,y^'(0)=1 ?
The solution for y^{''}+2y^'-8y=0,y(0)=-2,y^'(0)=1 is y=(-2+5/6)e^{2t}-5/6 e^{-4t}