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

Solution steps

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

Solve linear ODE:

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

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

(1 - cos (t))^{^{'}}

\int_{0}^{2} 1 + 16 x^{2} d x

\int \frac{1}{x ^{15}} d x

x \to π lim (5 \frac{x - π}{x + π})

\frac{d y}{d x} + 4 y = 60

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