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}e^{-2t}+c_{3}e^{2t}

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

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

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

\int_{0}^{6} (6 x + 2) d x

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

x \to 2 lim (F (x))

\frac{d}{d x} ((6 x^{2} + 7) (6 x + 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}e^{-2t}+c_{3}e^{2t}