What is the solution for y^{'''}+y^{''}+y^'-39y=0 ?

The solution for y^{'''}+y^{''}+y^'-39y=0 is y=c_{1}e^{3t}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}+y^{''}+y^'-39y=0

y^{^{'''}} + y^{^{''}} + y^{^{'}} - 39 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) + y^{'} (t) - 39 y = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 3, y^{^{'}} (0) = 16

y^{^{''}} + 4 y + 68 y = 0, y (0) = 7

3 y^{^{''}} + 48 y = 0, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''''}} + 4 y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} - 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + \frac{df}{d x} = 0

What is the solution for y^{'''}+y^{''}+y^'-39y=0 ?
The solution for y^{'''}+y^{''}+y^'-39y=0 is y=c_{1}e^{3t}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))