What is the solution for y^{'''}-y^{''}+y^'-y=sec(t) ?

The solution for y^{'''}-y^{''}+y^'-y=sec(t) is

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y^{'''}-y^{''}+y^'-y=sec(t)

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = sec (t)

y = c_{1} e^{t} + c_{2} cos (t) + c_{3} sin (t) + e^{t} \cdot \int \frac{sec ( t )}{2 e ^{t}} d t + \frac{1}{2} cos (t) (- t - ln (cos (t))) - \frac{1}{2} sin (t) (- ln (cos (t)) + t)

Solution steps

y^{'''} (t) - y^{''} (t) + y^{'} (t) - y = sec (t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} cos (t) + c_{3} sin (t) + e^{t} \cdot \int \frac{sec ( t )}{2 e ^{t}} d t + \frac{1}{2} cos (t) (- t - ln (cos (t))) - \frac{1}{2} sin (t) (- ln (cos (t)) + t)

y = c_{1} e^{t} + c_{2} cos (t) + c_{3} sin (t) + e^{t} \cdot \int \frac{sec ( t )}{2 e ^{t}} d t + \frac{1}{2} cos (t) (- t - ln (cos (t))) - \frac{1}{2} sin (t) (- ln (cos (t)) + t)

y^{^{''}} + y = cos (x) - sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 1, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 4 x

y^{^{''''}} - y = e^{2 t}

- \frac{1}{9} y^{^{''}} + \frac{2}{3} y^{^{'}} - y = \frac{t}{t + 1}

What is the solution for y^{'''}-y^{''}+y^'-y=sec(t) ?
The solution for y^{'''}-y^{''}+y^'-y=sec(t) is