What is the solution for y^{'''}+2y^{''}+5y^'-26y=0 ?

The solution for y^{'''}+2y^{''}+5y^'-26y=0 is y=c_{1}e^{2t}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}+2y^{''}+5y^'-26y=0

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) + 5 y^{'} (t) - 26 y = 0

Solve linear ODE:

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

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

y = \frac{x ^{2}}{2} + 4 x + 2, (- 3, - \frac{11}{2})

y^{^{'}} + \frac{1}{2} y = y^{3}

\frac{1}{2} e^{- t} cos (t) + \frac{1}{2} e^{- t} sin (t) - \frac{1}{2} e^{- 2 t}

\frac{d}{d x} (cos^{2} (ln (x)))

y^{^{''}} + 8 y^{^{'}} + 6 y = 0

What is the solution for y^{'''}+2y^{''}+5y^'-26y=0 ?
The solution for y^{'''}+2y^{''}+5y^'-26y=0 is y=c_{1}e^{2t}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))