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

The solution for (dr)/(dθ)+rtan(θ)=sec(θ)tan(θ) is r(θ)=(tan^2(θ))/(2sec(θ))+(c_{1})/(sec(θ))

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(dr)/(dθ)+rtan(θ)=sec(θ)tan(θ)

\frac{d r}{d θ} + r tan (θ) = sec (θ) tan (θ)

r (θ) = \frac{tan ^{2} ( θ )}{2 sec ( θ )} + \frac{c _{1}}{sec ( θ )}

Solution steps

\frac{d r}{d θ} + r tan (θ) = sec (θ) tan (θ)

Solve linear ODE:

r (θ) = \frac{tan ^{2} ( θ )}{2 sec ( θ )} + \frac{c _{1}}{sec ( θ )}

r (θ) = \frac{tan ^{2} ( θ )}{2 sec ( θ )} + \frac{c _{1}}{sec ( θ )}

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

f (x) = x^{2} + 3 x^{6}

\frac{\partial}{\partial x} (e^{2 y})

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

x \to 1 lim (- x^{2} + 5 x - 7)

What is the solution for (dr)/(dθ)+rtan(θ)=sec(θ)tan(θ) ?
The solution for (dr)/(dθ)+rtan(θ)=sec(θ)tan(θ) is r(θ)=(tan^2(θ))/(2sec(θ))+(c_{1})/(sec(θ))