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

The solution for linear y^{''}-y^'-2y=0,y(0)=1,y^'(0)=-1 is y=e^{-t}

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

linear

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

y = e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{- t}

y = e^{- t}

y^{^{''}} - 2 y^{^{'}} - 10 y = 0

0.5 y^{^{''}} + y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + \frac{a}{4} y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x (0) = 1, x^{^{'}} (0) = 0

4 y^{^{''}} + \frac{16}{0.25} y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for linear y^{''}-y^'-2y=0,y(0)=1,y^'(0)=-1 ?
The solution for linear y^{''}-y^'-2y=0,y(0)=1,y^'(0)=-1 is y=e^{-t}