What is the solution for y^{'''}+4y^'=sec(x)cos(x) ?

The solution for y^{'''}+4y^'=sec(x)cos(x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+x/4

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y^{'''}+4y^'=sec(x)cos(x)

y^{^{'''}} + 4 y^{^{'}} = sec (x) cos (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{x}{4}

Solution steps

y^{'''} (x) + 4 y^{'} (x) = sec (x) cos (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{x}{4}

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{x}{4}

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = 5 + 5 x

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 15 t

y^{^{''}} + 9 y = e^{x}

y^{^{''''}} + 8 y^{^{''}} + 16 y = 48 + cos (4 t)

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