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

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

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

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

y = c_{1} e^{t} + c_{2} cos (6 t) + c_{3} sin (6 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} cos (6 t) + c_{3} sin (6 t)

y = c_{1} e^{t} + c_{2} cos (6 t) + c_{3} sin (6 t)

\frac{d y}{d x} = \frac{3 y}{x ^{2}}

(t + 2 y + 5) d t - d y = 0

\frac{d}{d x} ((\frac{3 x ^{2}}{2 x + 4})^{3})

\frac{5 \cdot e ^{- 2 s}}{s - 1}

\int cos (x) + sec^{2} (x) d x

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