What is the solution for y^{'''}+16y^'=e^{-pix} ?

The solution for y^{'''}+16y^'=e^{-pix} is y=c_{1}+c_{2}cos(4x)+c_{3}sin(4x)-1/(pi^3+16pi)e^{-pix}

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y^{'''}+16y^'=e^{-pix}

y^{^{'''}} + 16 y^{^{'}} = e^{- π x}

y = c_{1} + c_{2} cos (4 x) + c_{3} sin (4 x) - \frac{1}{π ^{3} + 16 π} e^{- π x}

Solution steps

y^{'''} (x) + 16 y^{'} (x) = e^{- π x}

Solve linear ODE:

y = c_{1} + c_{2} cos (4 x) + c_{3} sin (4 x) - \frac{1}{π ^{3} + 16 π} e^{- π x}

y = c_{1} + c_{2} cos (4 x) + c_{3} sin (4 x) - \frac{1}{π ^{3} + 16 π} e^{- π x}

x^{^{''}} + 4 x = 4 sec^{2} (2 t)

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x} cos (2 x)

y^{^{''}} + y = cos (- t + \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} - 2 y = x

What is the solution for y^{'''}+16y^'=e^{-pix} ?
The solution for y^{'''}+16y^'=e^{-pix} is y=c_{1}+c_{2}cos(4x)+c_{3}sin(4x)-1/(pi^3+16pi)e^{-pix}