What is the solution for (d^2+1)y=tan(x) ?

The solution for (d^2+1)y=tan(x) is y=c_{1}cos(x)+c_{2}sin(x)-cos(x)ln(tan(x)+sec(x))

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(d^2+1)y=tan(x)

(d^{2} + 1) y = tan (x)

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

Solution steps

(d^{2} + 1) y = tan (x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

(d^{2} - 3 d + 2) \cdot y = 3 sin (x)

y^{^{'''}} - 4 y^{^{'}} = 0

y^{^{'''}} - 4 y = 0

2 y^{^{''}} - 3 y^{^{'}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

2 (d^{4} + 11 d^{3} - 4 d^{2} - 69 d + 34) y = 0

What is the solution for (d^2+1)y=tan(x) ?
The solution for (d^2+1)y=tan(x) is y=c_{1}cos(x)+c_{2}sin(x)-cos(x)ln(tan(x)+sec(x))