What is the solution for y^{'''}+36y^'=0,y^{''}(0)=108 ?

The solution for y^{'''}+36y^'=0,y^{''}(0)=108 is y=c_{1}-3cos(6t)+c_{3}sin(6t)

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y^{'''}+36y^'=0,y^{''}(0)=108

y^{^{'''}} + 36 y^{^{'}} = 0, y^{^{''}} (0) = 108

y = c_{1} - 3 cos (6 t) + c_{3} sin (6 t)

Solution steps

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

Solve linear ODE:

y = c_{1} - 3 cos (6 t) + c_{3} sin (6 t)

y = c_{1} - 3 cos (6 t) + c_{3} sin (6 t)

x^{^{''}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = - 22

y^{^{''}} = 3 y^{^{'}} + 5 y, y (2) = 8

y^{^{''}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y^{^{''}} - y^{^{'}} - 72 y = 0, y (1) = e^{- 8 x}

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

What is the solution for y^{'''}+36y^'=0,y^{''}(0)=108 ?
The solution for y^{'''}+36y^'=0,y^{''}(0)=108 is y=c_{1}-3cos(6t)+c_{3}sin(6t)