What is the solution for y^{'''}+3y^{''}-18y^'-40y=0 ?

The solution for y^{'''}+3y^{''}-18y^'-40y=0 is y=c_{1}e^{-2t}+c_{2}e^{4t}+c_{3}e^{-5t}

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y^{'''}+3y^{''}-18y^'-40y=0

y^{^{'''}} + 3 y^{^{''}} - 18 y^{^{'}} - 40 y = 0

y = c_{1} e^{- 2 t} + c_{2} e^{4 t} + c_{3} e^{- 5 t}

Solution steps

y^{'''} (t) + 3 y^{''} (t) - 18 y^{'} (t) - 40 y = 0

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{4 t} + c_{3} e^{- 5 t}

y = c_{1} e^{- 2 t} + c_{2} e^{4 t} + c_{3} e^{- 5 t}

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

2 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 105 y = 0

6 y^{^{''''}} + 77 y^{^{''}} + 255 y = 0

y^{^{''}} - 0.28 y^{^{'}} - 0.006 y = 0

2 \cdot y^{^{''}} + 4 \cdot y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for y^{'''}+3y^{''}-18y^'-40y=0 ?
The solution for y^{'''}+3y^{''}-18y^'-40y=0 is y=c_{1}e^{-2t}+c_{2}e^{4t}+c_{3}e^{-5t}