Question

please help me.i will make you brainlist ​

Answer 1

Answer:

(x+tany)dy=sin2ydx

⇒

dy

dx

=

sin2y

x+tany

⇒

dy

dx

=

sin2y

x

+

sin2y

tany

⇒

dy

dx

−cosec2y.x=

2

1

sec

2

y(∵sinx=

cosecx

1

;tanx=

cosx

sinx

;sin2x=2sinxcosx)

sincethisisintheformof

dy

dx

+Rx=S

where,R=−cosec2yandS=

2

1

sec

2

y

∴I.F=e

∫Rdy

=e

∫(−cosec2y)dy

=e

−log∣cosec2y−cot2y∣

=e

−log(tany)

=e

log(coty)

=coty

sothesolutionofthegivendifferentialequationisgivenby

x.(I.F)=∫S.(I.F)dy+C

⇒x.(coty)=∫

2

1

sec

2

y.(coty)dy+C

⇒x.(coty)=∫

sin2y

1

dy+C

⇒x.(coty)=∫cosec2ydy+C

⇒x.(coty)=

2

1

log∣cos2y−cot2y∣+C