What is the solution for y^{'''}-2y^'-4y=0 ?

The solution for y^{'''}-2y^'-4y=0 is y=c_{1}e^{2t}+e^{-t}(c_{2}cos(t)+c_{3}sin(t))

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y^{'''}-2y^'-4y=0

y^{^{'''}} - 2 y^{^{'}} - 4 y = 0

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

Solution steps

y^{'''} (t) - 2 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

t + 1^{2}

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What is the solution for y^{'''}-2y^'-4y=0 ?
The solution for y^{'''}-2y^'-4y=0 is y=c_{1}e^{2t}+e^{-t}(c_{2}cos(t)+c_{3}sin(t))