What is the solution for y^{'''}+36y^'=0,y(0)=1 ?

The solution for y^{'''}+36y^'=0,y(0)=1 is y=1-c_{2}+c_{2}cos(6t)+c_{3}sin(6t)

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y^{'''}+36y^'=0,y(0)=1

y^{^{'''}} + 36 y^{^{'}} = 0, y (0) = 1

y = 1 - c_{2} + c_{2} cos (6 t) + c_{3} sin (6 t)

Solution steps

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

Solve linear ODE:

y = 1 - c_{2} + c_{2} cos (6 t) + c_{3} sin (6 t)

y = 1 - c_{2} + c_{2} cos (6 t) + c_{3} sin (6 t)

y^{^{''}} - y^{^{'}} = - 10 y

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

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

y^{^{''}} - 5 y^{^{'}} - 36 y = 0, y^{0} = a, y^{^{'}} 0 = 32

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos^{2} (x)

What is the solution for y^{'''}+36y^'=0,y(0)=1 ?
The solution for y^{'''}+36y^'=0,y(0)=1 is y=1-c_{2}+c_{2}cos(6t)+c_{3}sin(6t)