Question

Solve : x.cosy. dx = y.cosx. dy​

Answer 1

Answer:

xcos

x

y

+ysin

x

y

]y=x[ysin

x

y

−xcos

x

y

]×

dx

dy

Given differential eqn can be written as

dx

dy

=

x[ysin

x

y

−x cos

x

y

]

y[xcos

x

y

+ysin

x

y

]

....(1)

Substitute y=vx

dx

dy

=v+x

dx

dv

So eqn (1) becomes

v+x

dx

dv

=

[vsinv−cosv]

v[cosv+vsinv]

x

dx

dv

=

vsinv−cosv

2vcosv

vcosv

vsinv−cosv

dv=

x

2

dx

Integrating both sides,

∫tanvdv−∫

v

1

dv=2∫

x

1

dx

−logcosv−logv=2logx+logC

⇒

vcosv

1

=Cx

2

⇒xycos(

x

y

)=

C

1

⇒xycos(

x

y

)=c