What is the solution for y^{'''}+16y^'=0 ?

The solution for y^{'''}+16y^'=0 is y=c_{1}+c_{2}cos(4t)+c_{3}sin(4t)

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

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

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

Solution steps

y^{'''} (t) + 16 y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{'}} = \frac{( x - 1 )}{y}

\frac{d}{d x} (\frac{1}{3 ^{(- x^{2} - y^{2})} + 1})

\int 3 t^{- 4} (2 + 4 t^{- 3})^{- 7} d t

\int t^{3} t^{4} + 5 d t

What is the solution for y^{'''}+16y^'=0 ?
The solution for y^{'''}+16y^'=0 is y=c_{1}+c_{2}cos(4t)+c_{3}sin(4t)