Question

Solution of the differential equation sec^2 x tan⁡y dx + sec^2 y tan⁡x dy=0  is _______

(a) (sec x. sec y)=k

(b) (sec x .tany)=k

(c) (tan x. tany)=k

(d) (sec x .tan x)+(sec y .tan y)=k

I got this question during a job interview.

I would like to ask this question from Separable and Homogeneous Equations in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

The correct option is (c) (tan x. tany)=k

The best explanation: sec^2 x tan⁡y dx + sec^2 y tan⁡x dy=0

dividing throughout by tan y.tan x we get

\(\frac{sec^2 x}{tan⁡x}  \,dx + \frac{sec^2 y}{tan⁡y} \,dy=0\)……separating the variable

now integrating we get \(\int \frac{sec^2 x}{tan⁡x}  \,dx + \int \frac{sec^2 y}{tan⁡y} \,dy=c\)

 substituting tan x = t & tan y=p→sec^2 x dx=dt & sec^2 y dy=dp

\( \rightarrow \int \frac{1}{t} \,dt + \int \frac{1}{p} \,dp=c\)

log t + log p = c –>log(tan x)+log(tan y) = c = log k….since it is an unknown constant

log(tan x .tan y) = log k

(tan x tan y) = k is the solution.