What is the solution for (d^2-2d+2)y=e^xtan(x) ?

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

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(d^2-2d+2)y=e^xtan(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 6 y = 11

(d + 2) (d - 1)^{3} y = e^{x}

(d^{4} - 1) y = e^{x} cos (x) cos^{3} (x)

(d^{2} - 4 d + 4) \cdot y = 8 e^{2 x} \cdot x^{2} \cdot sin^{2} (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t - 8

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