What is the solution for y^{'''}+5y^{''}-4y^'-60y=0 ?

The solution for y^{'''}+5y^{''}-4y^'-60y=0 is y=c_{1}e^{3t}+e^{-4t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+5y^{''}-4y^'-60y=0

y^{^{'''}} + 5 y^{^{''}} - 4 y^{^{'}} - 60 y = 0

y = c_{1} e^{3 t} + e^{- 4 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

Solution steps

y^{'''} (t) + 5 y^{''} (t) - 4 y^{'} (t) - 60 y = 0

Solve linear ODE:

y = c_{1} e^{3 t} + e^{- 4 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

y = c_{1} e^{3 t} + e^{- 4 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

25 y^{^{''}} + 10 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 10 y^{^{'}} + 26 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

x^{^{'''}} + 4 x^{^{''}} + 13 x^{^{'}} = 0

(d^{4} - 5 d^{3} + 5 d^{2} + 5 d - 6) y = 0

What is the solution for y^{'''}+5y^{''}-4y^'-60y=0 ?
The solution for y^{'''}+5y^{''}-4y^'-60y=0 is y=c_{1}e^{3t}+e^{-4t}(c_{2}cos(2t)+c_{3}sin(2t))