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

##### 1 Answer

# sec^2y = sec^2x + C #

#### Explanation:

We have:

# cosy(ln(secx+tanx))dx=cosx(ln(secy+tany))dy # ..... [A]

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

# (ln(secy+tany))/Cosydy/dx = (ln(secx+tanx))/ cosx #

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

# int \ (ln(secy+tany))/Cosy \ dy = int \ (ln(secx+tanx))/ cosx \ dx # ..... [B}

Consider the RHS integral:

# I = int \ (ln(secx+tanx))/ cosx \ dx #

# \ \ = int \ secx \ (ln(secx+tanx)) \ dx #

We can perform a substitution:

# u = sec x => (du)/dx = ln|secx+tanx|#

if we substitute this into the integral we get:

# I = int \ u \ du = 1/2u^2+A #

# \ \ = 1/2sec^2x + A #

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

# 1/2sec^2y = 1/2sec^2x + A #

# :. sec^2y = sec^2x + C #

Which is the General Solution of [A]