What is the solution for y^{''}-2y^'-8y=0,y(0)=α,y^'(0)=5 ?

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

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y^{''}-2y^'-8y=0,y(0)=α,y^'(0)=5

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = α, y^{^{'}} (0) = 5

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 61 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} + 4 y^{^{''}} - 28 y^{^{'}} + 32 y = 0

y^{^{'''}} - 9 y^{^{''}} = 0

y^{^{''}} - 5 y^{^{'}} + 7 y = 0

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