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

The solution for y^{''}-y^'-2y=0,y(0)=α,y^'(0)=2 is y=e^{2t}(α+(2-2α)/3)-(e^{-t}(2-2α))/3

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y^{''}-y^'-2y=0,y(0)=α,y^'(0)=2

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

y = e^{2 t} (α + \frac{2 - 2 α}{3}) - \frac{e ^{- t} ( 2 - 2 α )}{3}

Solution steps

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

Solve linear ODE:

y = e^{2 t} (α + \frac{2 - 2 α}{3}) - \frac{e ^{- t} ( 2 - 2 α )}{3}

y = e^{2 t} (α + \frac{2 - 2 α}{3}) - \frac{e ^{- t} ( 2 - 2 α )}{3}

y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d t ^{2}} + 2 y = 0, y^{^{'}} (0) = 2

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

\frac{1}{2} y^{^{''}} + 2 y = 0

y^{^{'''}} + 6 y^{^{''}} = 0

What is the solution for y^{''}-y^'-2y=0,y(0)=α,y^'(0)=2 ?
The solution for y^{''}-y^'-2y=0,y(0)=α,y^'(0)=2 is y=e^{2t}(α+(2-2α)/3)-(e^{-t}(2-2α))/3