What is the solution for y^{''}-y^'-2*y=0 ?

The solution for y^{''}-y^'-2*y=0 is y=c_{1}e^{2t}+c_{2}e^{-t}

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

y^{^{''}} - y^{^{'}} - 2 \cdot y = 0

y = c_{1} e^{2 t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t}

y = c_{1} e^{2 t} + c_{2} e^{- t}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 68 y = 0, y (0) = 7

2 x^{^{''}} + 32 x = 0, x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} + y = 0, y (0) = 11, y (0) = 33

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

What is the solution for y^{''}-y^'-2*y=0 ?
The solution for y^{''}-y^'-2*y=0 is y=c_{1}e^{2t}+c_{2}e^{-t}