Question

solve (sec x tan x tan y -e^x)dx+ sec x sec^2y dy =0

Answer 1

Answer:

secx⋅tanx⋅tany−ex)⋅dx+(secx⋅sec2y)⋅dy=0

⟹secx⋅tanx⋅tany⋅dx+secx⋅sec2y⋅dy=ex⋅dx

⟹d(secx⋅tany)=d(ex)

⟹secx⋅tany=ex+C

Step-by-step explanation:

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Answer 2

Answer:

Solving (sec x tan x tan y -eˣ)dx+ sec x sec²y dy =0

gives sec x tan x = eˣ + c    

Step-by-step explanation:

Given (sec x tan x tan y -eˣ)dx+ sec x sec²y dy =0

This can be written as

(sec x tan x tan y)dx - (eˣ)dx + (sec x sec²y)dy =0

=> (sec x tan x tan y)dx + (sec x sec²y)dy =  (eˣ)dx

Now we know that d(sec x * tan y) = sec x sec²y + tan y *sec x * tan x

On substitution,

d(sec x tan x) = (eˣ)dx

On integrating both sides, we get,

$\int$ d(sec x tan x) = $\int$ (eˣ)dx

=> sec x tan x = eˣ + c                

which is the required solution.