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

The solution for y^{'''}+25y^'=0 is y=c_{1}+c_{2}cos(5t)+c_{3}sin(5t)

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

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

y = c_{1} + c_{2} cos (5 t) + c_{3} sin (5 t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (5 t) + c_{3} sin (5 t)

y = c_{1} + c_{2} cos (5 t) + c_{3} sin (5 t)

y^{^{'''}} + 2 y^{^{''}} - 20 y^{^{'}} + 24 y = 0

0.25 y^{^{''}} + 10 y^{^{'}} + 1000 y = 0

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

y^{^{''}} - 8 y^{^{''}} + 16 y = 0

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

What is the solution for y^{'''}+25y^'=0 ?
The solution for y^{'''}+25y^'=0 is y=c_{1}+c_{2}cos(5t)+c_{3}sin(5t)