What is the solution for y^{'''}+8y^{''}+25y^'=0 ?

The solution for y^{'''}+8y^{''}+25y^'=0 is y=c_{1}+e^{-4t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}+8y^{''}+25y^'=0

y^{^{'''}} + 8 y^{^{''}} + 25 y^{^{'}} = 0

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

Solution steps

y^{'''} (t) + 8 y^{''} (t) + 25 y^{'} (t) = 0

Solve linear ODE:

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

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

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

x^{^{'''}} - 10 x^{^{''}} + 37 x^{^{'}} - 52 x = 0

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 7

y^{^{''}} + 8 y^{^{'}} + 32 y = 0, y^{^{'}} 0 = - 4, y^{0} = 1

4 y^{^{''}} + 3 y^{^{''}} + y = 0

What is the solution for y^{'''}+8y^{''}+25y^'=0 ?
The solution for y^{'''}+8y^{''}+25y^'=0 is y=c_{1}+e^{-4t}(c_{2}cos(3t)+c_{3}sin(3t))