What is the solution for y^'-ytan(x)=cos(x) ?

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

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y^'-ytan(x)=cos(x)

y^{^{'}} - y tan (x) = cos (x)

y = \frac{2 x + sin ( 2 x )}{4 cos ( x )} + \frac{c _{1}}{cos ( x )}

Solution steps

y^{'} (x) - y tan (x) = cos (x)

Solve linear ODE:

y = \frac{2 x + sin ( 2 x )}{4 cos ( x )} + \frac{c _{1}}{cos ( x )}

y = \frac{2 x + sin ( 2 x )}{4 cos ( x )} + \frac{c _{1}}{cos ( x )}

\frac{\partial}{\partial s} (sin (s + t))

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

\frac{2}{5} x^{2} + \frac{1}{4} x - 1

\frac{d}{d x} (4 x^{2} + 6 x)

8 x^{4} - x^{2} + 9 x + 2

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