What is the general solution of the differential equation? $cos y (ln (sec x + tan x)) d x = cos x (ln (sec y + tan y)) d y$ $cosy(ln(secx+tanx))dx=cosx(ln(secy+tany))dy$

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Answer 1 · 2017-10-05T10:13:54

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

We have:

$cos y (ln (sec x + tan x)) d x = cos x (ln (sec y + tan y)) d y$ $cosy(ln(secx+tanx))dx=cosx(ln(secy+tany))dy$ ..... [A]

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

$\frac{ln (sec y + tan y)}{cos y} \frac{d y}{d x} = \frac{ln (sec x + tan x)}{cos x}$ $(ln(secy+tany))/Cosydy/dx = (ln(secx+tanx))/ cosx$

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]

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