What is the solution for y^{'''}+2y^{''}-15y^'-36y=0 ?

The solution for y^{'''}+2y^{''}-15y^'-36y=0 is y=c_{1}e^{4t}+c_{2}e^{-3t}+c_{3}te^{-3t}

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y^{'''}+2y^{''}-15y^'-36y=0

y^{^{'''}} + 2 y^{^{''}} - 15 y^{^{'}} - 36 y = 0

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) - 15 y^{'} (t) - 36 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 6 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

2 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

y^{^{'''}} + 8 y^{^{'}} = 0

What is the solution for y^{'''}+2y^{''}-15y^'-36y=0 ?
The solution for y^{'''}+2y^{''}-15y^'-36y=0 is y=c_{1}e^{4t}+c_{2}e^{-3t}+c_{3}te^{-3t}