What is the solution for (dr)/(dθ)+rtan(θ)=sec(θ) ?

The solution for (dr)/(dθ)+rtan(θ)=sec(θ) is r(θ)=(tan(θ))/(sec(θ))+(c_{1})/(sec(θ))

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(dr)/(dθ)+rtan(θ)=sec(θ)

\frac{d r}{d θ} + r tan (θ) = sec (θ)

r (θ) = \frac{tan ( θ )}{sec ( θ )} + \frac{c _{1}}{sec ( θ )}

Solution steps

\frac{d r}{d θ} + r tan (θ) = sec (θ)

Solve linear ODE:

r (θ) = \frac{tan ( θ )}{sec ( θ )} + \frac{c _{1}}{sec ( θ )}

r (θ) = \frac{tan ( θ )}{sec ( θ )} + \frac{c _{1}}{sec ( θ )}

f (x) = 6 sin (x) - 4 x cos (x) - x^{2} sin (x)

\frac{\partial}{\partial y} (50 y + 3000 - 100 x)

\frac{\partial}{\partial x} (\frac{59}{( 1 + x ^{2} + y ^{2} )})

(1 - x - x^{2}) (x - 2), at x = 1

(1 + x^{2}) y^{^{'}} + 4 x y = x, y (0) = 1

What is the solution for (dr)/(dθ)+rtan(θ)=sec(θ) ?
The solution for (dr)/(dθ)+rtan(θ)=sec(θ) is r(θ)=(tan(θ))/(sec(θ))+(c_{1})/(sec(θ))