# What is the general solution of the differential equation?  cosy(ln(secx+tanx))dx=cosx(ln(secy+tany))dy

Oct 5, 2017

${\sec}^{2} y = {\sec}^{2} x + C$

#### Explanation:

We have:

$\cos y \left(\ln \left(\sec x + \tan x\right)\right) \mathrm{dx} = \cos x \left(\ln \left(\sec y + \tan y\right)\right) \mathrm{dy}$ ..... [A]

If we rearrange this ODE from differential form into standard form we have:

$\frac{\ln \left(\sec y + \tan y\right)}{C} o s y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\ln \left(\sec x + \tan x\right)}{\cos} x$

This is now a separable ODE, do we can "separate the variables" to give:

$\int \setminus \frac{\ln \left(\sec y + \tan y\right)}{C} o s y \setminus \mathrm{dy} = \int \setminus \frac{\ln \left(\sec x + \tan x\right)}{\cos} x \setminus \mathrm{dx}$ ..... [B}

Consider the RHS integral:

$I = \int \setminus \frac{\ln \left(\sec x + \tan x\right)}{\cos} x \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \sec x \setminus \left(\ln \left(\sec x + \tan x\right)\right) \setminus \mathrm{dx}$

We can perform a substitution:

$u = \sec x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \ln | \sec x + \tan x |$

if we substitute this into the integral we get:

$I = \int \setminus u \setminus \mathrm{du} = \frac{1}{2} {u}^{2} + A$
$\setminus \setminus = \frac{1}{2} {\sec}^{2} x + A$

Using this result we can now integrate both sides of [B] to get:

$\frac{1}{2} {\sec}^{2} y = \frac{1}{2} {\sec}^{2} x + A$

$\therefore {\sec}^{2} y = {\sec}^{2} x + C$

Which is the General Solution of [A]