What is the solution for y^{'''}+2y^{''}-11y^'-52y=0 ?

The solution for y^{'''}+2y^{''}-11y^'-52y=0 is y=c_{1}e^{4t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+2y^{''}-11y^'-52y=0

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 52 y = 0

y = c_{1} e^{4 t} + e^{- 3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

Solution steps

y^{'''} (t) + 2 y^{''} (t) - 11 y^{'} (t) - 52 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + e^{- 3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

y = c_{1} e^{4 t} + e^{- 3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

\frac{d ^{2} f}{d x ^{2}} + 64 f (x) = 0

5 y^{^{''}} - 7 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 0.8 y

(D^{3} - D^{2} - 8 D + 12) y = 0

2 y^{^{''}} - 5 y^{^{'}} - 7 y = 0, y^{^{'}} (0) = 3, y (0) = 4

f^{^{''}} (t) + f^{^{'}} (t) + \frac{1}{16} f (t) = 0

What is the solution for y^{'''}+2y^{''}-11y^'-52y=0 ?
The solution for y^{'''}+2y^{''}-11y^'-52y=0 is y=c_{1}e^{4t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))