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

The solution for y^{'''}+4y^'=tan(x) is

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y^{'''}+4y^'=tan(x)

y^{^{'''}} + 4 y^{^{'}} = tan (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{1}{4} ln (cos (x)) - \frac{cos ( 2 x ) ( tan ( x ) sin ( 2 x ) + ln ( cos ^{2} ( x ) ) )}{8} - \frac{1}{4} sin (2 x) (x - \frac{1}{2} sin (2 x))

Solution steps

y^{'''} (x) + 4 y^{'} (x) = tan (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{1}{4} ln (cos (x)) - \frac{cos ( 2 x ) ( tan ( x ) sin ( 2 x ) + ln ( cos ^{2} ( x ) ) )}{8} - \frac{1}{4} sin (2 x) (x - \frac{1}{2} sin (2 x))

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{1}{4} ln (cos (x)) - \frac{cos ( 2 x ) ( tan ( x ) sin ( 2 x ) + ln ( cos ^{2} ( x ) ) )}{8} - \frac{1}{4} sin (2 x) (x - \frac{1}{2} sin (2 x))

y^{^{''}} - 10 y^{^{'}} + 25 y = 25 x + 10

y^{^{''}} - 7 y^{^{'}} + 12 = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 42 x + 5

d (x) = 5 (x - 9)^{2} - 32

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

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