What is the solution for y^{'''}+y^{''}+3y^'-5y=0 ?

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

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y^{'''}+y^{''}+3y^'-5y=0

y^{^{'''}} + y^{^{''}} + 3 y^{^{'}} - 5 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) + 3 y^{'} (t) - 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 24 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 22

y^{^{''}} - y = 0, y^{^{'}} 0 = 6, y^{0} (0) = 2

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{d ^{2} x}{d t ^{2}} + c \cdot \frac{d x}{d t} + k \cdot x = 0

y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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