What is the solution for y^{'''}-y^{''}-100y=0 ?

The solution for y^{'''}-y^{''}-100y=0 is y=c_{1}e^{5t}+e^{-2t}(c_{2}cos(4t)+c_{3}sin(4t))

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y^{'''}-y^{''}-100y=0

y^{^{'''}} - y^{^{''}} - 100 y = 0

y = c_{1} e^{5 t} + e^{- 2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

Solution steps

y^{'''} (t) - y^{''} (t) - 100 y = 0

Solve linear ODE:

y = c_{1} e^{5 t} + e^{- 2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

y = c_{1} e^{5 t} + e^{- 2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

25 y^{^{''}} + 50 y^{^{'}} + 169 y = 0, y (0) = 4, y^{^{'}} (0) = 8

64 y^{^{''}} + 80 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} + \frac{0}{3} y^{^{'}} + \frac{12}{3} y = 0

3 y^{^{'''}} - 2 y^{^{''}} - 37 y^{^{'}} - 12 y = 0

What is the solution for y^{'''}-y^{''}-100y=0 ?
The solution for y^{'''}-y^{''}-100y=0 is y=c_{1}e^{5t}+e^{-2t}(c_{2}cos(4t)+c_{3}sin(4t))