Question

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

Answer 1

`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]