Question

the solution of differential equations sec2 y ​

Answer 1

Answer:

iven, sec  

2

x⋅tanydx+sec  

2

y⋅tanxdy=0

On separating the varaibales, we get

⇒sec  

2

x⋅tanydx=−sec  

2

y⋅tanxdy

⇒  

tanx

sec  

2

x

​  

dx=−  

tany

sec  

2

y

​  

dy

On integration both the sides, we get

∫  

tanx

sec  

2

x

​  

dx=−∫  

tany

sec  

2

y

​  

dy

Let tanx=u⇒sec  

2

x=  

dx

du

​  

 

⇒dx=  

sec  

2

x

du

​  

 and tany=v

⇒sec  

2

y=  

dy

dv

​  

 

⇒dy=  

sec  

2

y

dv

​  

 

∴∫  

u

sec  

2

x

​  

⋅  

sec  

2

x

du

​  

=−∫  

v

sec  

2

y

​  

⋅  

sec  

2

y

dv

​  

 

⇒∫  

u

du

​  

=∫  

v

dv

​  

 

⇒log∣u∣=−log∣v∣+log∣C∣

⇒log∣tanx∣=−log∣tany∣+log∣C∣

⇒log∣tanx⋅tany∣=log∣C∣

(∵logm+logn=logmn)

(∵

Step-by-step explanation: