Question

10.er tan y dr + (1-e") sec2 y dy = 0

Answer 1

The given differential equation is,

e x tanydx+( 1− e x ) sec 2 ydy=0

Separate the variables.

( 1− e x ) sec 2 ydy=− e x tanydx sec 2 y tany dy= − e x ( 1− e x ) dx

Integrating both the sides, we get,

∫ sec 2 y tany dy = ∫ − e x ( 1− e x ) dx (1)

Let,

I 1 = ∫ sec 2 y tany dy I 2 = ∫ − e x ( 1− e x ) dx

Let, tany=u.

Differentiate on both sides.

d dy ( tany )= du dy sec 2 y= du dy sec 2 ydy=du

Substitute the values in I 1 .

∫ sec 2 y tany dy = ∫ du u log| u |=log| tany | (2)

Let, 1− e x =t.

Differentiate on both the sides.

d dx ( 1− e x )= dt dx − e x = dt dx − e x dx=dt

Substitute the values in I 2 .

∫ − e x ( 1− e x ) dx = ∫ dt t =logt =log( 1− e −x ) (3)

Substitute the values of equations (2) and (3) in equation (1).

log| tany |=log( 1− e −x )+logC log| tany |=log| C( 1− e −x ) | tany=C( 1− e −x )

Thus, the general solution of the differential equation e x tanydx+( 1− e x ) sec 2 ydy=0 is given as tany=C( 1− e −x ).