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

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

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

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

y = - 2 e^{- t}

Solution steps

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

Solve linear ODE:

y = - 2 e^{- t}

y = - 2 e^{- t}

y^{^{''}} - 13 y^{^{'}} - 40 y = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y (0) = 3

x^{^{''}} = - 2 x^{^{'}}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (1) x = e^{3 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

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