What is the solution for y^{'''}+y^{''}-4y^'+6y=0 ?

The solution for y^{'''}+y^{''}-4y^'+6y=0 is y=c_{1}e^{-3t}+e^t(c_{2}cos(t)+c_{3}sin(t))

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y^{'''}+y^{''}-4y^'+6y=0

y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} + 6 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) - 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

14 y^{^{''}} - y^{^{'}} - 3 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - 9 y = 0

4 y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + a y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

y^{^{'''}} + 3 y^{^{''}} - 30 y^{^{'}} - 50 y = 0

What is the solution for y^{'''}+y^{''}-4y^'+6y=0 ?
The solution for y^{'''}+y^{''}-4y^'+6y=0 is y=c_{1}e^{-3t}+e^t(c_{2}cos(t)+c_{3}sin(t))