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

The solution for y^{''}-y^'-2y=0,y(0)=-2,y^'(0)=5 is y=e^{2t}-3e^{-t}

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

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

y = e^{2 t} - 3 e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{2 t} - 3 e^{- t}

y = e^{2 t} - 3 e^{- t}

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 8

2 x^{^{''}} + 4 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 7 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 4 y = 0

What is the solution for y^{''}-y^'-2y=0,y(0)=-2,y^'(0)=5 ?
The solution for y^{''}-y^'-2y=0,y(0)=-2,y^'(0)=5 is y=e^{2t}-3e^{-t}