What is the solution for y^{'''}+2y^{''}+y^'+2y=0 ?

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

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

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

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) + y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

n \to \infty lim (n^{- 2})

2, x, x^{2}

\frac{d}{d x} ((x^{3})^{\frac{1}{2}})

y = ln (x)

y = x^{2} - 4 x, y = - x^{2}

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