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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{x - 2}{x + 2})

x \to - 2 lim (\frac{∣ x + 2 ∣}{x})

\int e^{- 0.04 x} d x

x \to 500 lim (e^{x})

\int_{2}^{\infty} \frac{1}{x ^{2} + x} d x

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