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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 3 lim (\frac{4 x ^{2} - 36}{2 x - 6})

x \to 1 lim (3 x^{2} - 5 x + 4)

\int_{\frac{1}{2}}^{1} x^{- 3} - 11 d x

x^{^{''}} + 4 x^{^{'}} + 4 x = t^{- 2} e^{- 2 t}

\frac{d}{d x} (cos (x) - 5 tan (x))

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