Answer to Question #115394 in Differential Equations for Hakam

Question

Solve the differential equation
(tanxtany − e^ycosx) + y'sec^2 y = 0.

Answer 1

the question is wrong.

the question should be:

tanx tany - e^x cosx + y' sec²y = 0 equation1

solution: this equation is of the form:

f'(y) y' + P f(y) = Q where P and Q are functions of x.

here, P= tanx and Q = e^xcosx , f(y)=tany

let tany=v such that sec²y. y' = v'

equation1 becomes v'+ vtanx = e^xcosx equation2

this equation is linear in v and x

so, integrating factor= $\left(e^{\smash{\int P dx}}\right)$ = $\left(e^{\smash{\int tanx dx}}\right)$ = secx

equation2 becomes v.secx = $\int$ e^x cosx secs dx + constant'c'

v.secx = e^x + c

tany.secx = e^x +c