Question

The integrating factor of
logx dg/dx+y=2logy​

Answer 1

Answer:

xlogx

dx

dy

+y=2logx

⇒

dx

dy

+

xlogx

y

=

x

2

...(1)

Put P=

xlogx

1

⇒∫PdP=∫

xlogx

1

dx=log(logx)

∴I.F.=e

∫PdP

=e

log(logx)

=logx

Multiplying (1) by I.F. we get

logx

dx

dy

+

x

y

=

x

2

logx

Integrating both sides

ylogx=∫

x

2

logxdx+c=(logx)

2

+c

As y(e)=1

⇒1=1+c