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

The solution for 2y^{'''}-y^{''}-y^'=0 is y=c_{1}+c_{2}e^{-t/2}+c_{3}e^t

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

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

y = c_{1} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{t}

y = c_{1} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{t}

What is the solution for 2y^{'''}-y^{''}-y^'=0 ?
The solution for 2y^{'''}-y^{''}-y^'=0 is y=c_{1}+c_{2}e^{-t/2}+c_{3}e^t