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

The solution for y^{'''}+y^{''}-y^'-y=0 is y=c_{1}e^t+c_{2}e^{-t}+c_{3}te^{-t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{36}{( 1 + 3 x ) ^{3}})

(2, 2), (19, - 17)

\int tan^{3} (3 x) d x

\frac{d}{d x} (ln (x) - e^{- x})

2 x y d y = (x^{2} - y^{2}) d x

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