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

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

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x^{'''}+x^{''}-2x=0

x^{^{'''}} + x^{^{''}} - 2 x = 0

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

Solution steps

x^{'''} (t) + x^{''} (t) - 2 x = 0

Solve linear ODE:

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

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

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