What is the solution for 2y^{'''}-3y^{''}-18y^'-8y=0 ?

The solution for 2y^{'''}-3y^{''}-18y^'-8y=0 is y=c_{1}e^{-2t}+c_{2}e^{-t/2}+c_{3}e^{4t}

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2y^{'''}-3y^{''}-18y^'-8y=0

2 y^{^{'''}} - 3 y^{^{''}} - 18 y^{^{'}} - 8 y = 0

y = c_{1} e^{- 2 t} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{4 t}

Solution steps

2 y^{'''} (t) - 3 y^{''} (t) - 18 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{4 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- \frac{t}{2}} + c_{3} e^{4 t}

y = 3 x^{2} - 12, y = 0

\frac{d}{d x} (\frac{3 + sin ( x )}{3 x + cos ( x )})

y = e^{x}, y = e^{- 3 x}, x = - 2, x = 1

\frac{d}{d x} (e^{3 (x - 1)})

\int \frac{3}{2 + 3 x} d x

What is the solution for 2y^{'''}-3y^{''}-18y^'-8y=0 ?
The solution for 2y^{'''}-3y^{''}-18y^'-8y=0 is y=c_{1}e^{-2t}+c_{2}e^{-t/2}+c_{3}e^{4t}